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MATHS 5 Marks Questions

Probability · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 275 questions.

Questions · Page 2 of 6

5 Marks Questions

Question 515 Marks
From a lot of 10 bulbs, which includes 3 defectives, a sample of 2 bulbs is drawn at random. Find the probability distribution of the number of defective bulbs.
Answer
Let X be the random variate giving number of defective bulbs, X can take values 0, 1, 2 $\text{P(X = 0)}=\frac{\text{7c}_{2}}{\text{10c}_{2}}=\frac{7}{15},\text{P(X = 1)}=\frac{\text{7c}_{1}\times\text{3c}_{1}}{\text{10c}_{2}}=\frac{7}{15},\text{P(X = 2)}=\frac{\text{3c}_{2}}{\text{10c}_{2}}=\frac{1}{15}$ $\therefore$ Probability distribution of X is
X 0 1 2
P(X) 7/15 7/15 1/15
 
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Question 525 Marks
A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die.
Answer
Probability of success $( p) = \frac{1}{6},$ Prob. of failure $(q) = \frac{5}{6}.$Third six in sixth throw $\Rightarrow$ two successes in first five throws
$\therefore$ P(Two sixes in first five throws and third six in sixth throw)
$= 5_{C_{2}} \bigg(\frac{1}{6}\bigg)^{2}. \bigg(\frac{5}{6}\bigg)^{3} . \frac{1}{6}.$
$= 10 \frac{5^{3} 1}{6^{5}.6} = \frac{625}{23328}$
 
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Question 535 Marks
Three bags contain balls as shown in the table below:
Bag Number of White balls Number of Black balls Number of Red balls
I 1 2 3
II 2 1 1
III 4 3 2
A bag is chosen at random and two balls are drawn from it. They happen to be white and red. What is the probability that they came from the III bag?
Answer
Events are: $\text{E} _{1}:$ Choosing bag I$\text{E}_{2} :$ Choosing bag II
$\text{E}_{3} :$ Choosing bag III
$\text{A}:$ Getting a white and a red ball
$\therefore \text{P(E}_{1}) = \text{P(E}_{2}) = \text{P(E}_{3}) = \frac{1}{3}$
$\text{P}\bigg(\frac{\text{A}}{\text{E}_1}\bigg) = \frac{1.3}{6_{\text{c}_{2}}}=\frac{1}{5} ,\text{P}\bigg(\frac{\text{A}}{\text{E}_2}\bigg)\frac{2.1}{4_{\text{c}_{2}}} = \frac{1}{3},\text{P}\bigg(\frac{\text{A}}{\text{E}_3}\bigg) = \frac{4.2}{9_{\text{c}_{2}}} = \frac{2}{9}$
${P}\bigg(\frac{\text{E}_{3}}{\text{A}}\bigg) = \frac{P(E_{3)}P\bigg( \frac{A}{E_{3}}\bigg)}{\sum^{3}_{1} P (E_{1}).P\bigg(\frac{A}{E_{1}}\bigg)}$
$\frac{\frac{1}{3}.\frac{2}{9}}{\frac{1}{3}.\frac{1}{5} + \frac{1}{3} .\frac{1}{3} + \frac{1}{3} . \frac{1}{2}} = \frac{5}{17}$
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Question 545 Marks
A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of number of successes.
Answer
$\text{P (a doublet)} \frac{1}{6} \Rightarrow p = \frac{1}{6}, q = \frac{5}{6}$Probability distribution is given by $\bigg(\frac{1}{6} + \frac{5}{6}\bigg)^{4}$
Let X be the number of successes and P (X), the corresponding probability, where X takes values from 0 to 4
$\therefore$ The distribution is:
X 0 1 2 3 4
P (X) $\frac{625}{1296}$ $\frac{500}{1296}$ $\frac{150}{1296}$ $\frac{20}{1296}$ $\frac{1}{1296}$
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Question 555 Marks
An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accident involving a scooter, a car and a truck are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver.
Answer
$\text{Let E}_{1} ,\text{E}_{2}, \text{E}_{3}$ be the events of a person be a scooter driver, car driver and truck driver respectively.Let A be the event of a vehicle meeting an accident.
$\therefore \text{P} \text({E}_{1}) = \frac{1}{6}, \text{P} \text({E}_{2}) = \frac{1}{3}, \text{P} \text({E}_{3}) = \frac{1}{2}$
$\text{P} \bigg(\frac{A}{E_{1}}\bigg) = \frac{1}{100}, \text{P} \bigg(\frac{A}{E_{2}}\bigg) = \frac{3}{100}, \text{P} \bigg(\frac{A}{E_{3}}\bigg) = \frac{15}{100}$
$\text{P} \bigg(\frac{A}{E_{1}}\bigg) = \frac{P(E_{1}\big) \times P\bigg(\frac{A}{E_1}\bigg)}{\sum^{3}_ {i = 1} P (E_{i} \times \bigg(\frac{A}{E_{1}}\bigg)}, i = 1, 2, 3$
$= \frac{\frac{1}{6}\times \frac{1}{100}}{\frac{1}{6}\times\frac{1}{100} + \frac{1}{3} \times\frac{3}{100} + \frac{1}{2} \times\frac{15}{100}} \frac{\frac{1}{6}}{\frac{1}{6}+1+\frac{15}{2}} = \frac{1}{6}\times \frac{6}{52} = \frac{1}{52}$
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Question 565 Marks
Two numbers are selected at random (without replacement) from the first five positive integers. Let X denote the larger of the two numbers obtained. Find the mean and variance of X.
Answer
The first five positive integers are 1, 2, 3, 4, 5 we select two positive numbers in 5 × 4 = 20 ways.
Out of these two no. are selected at random.
let X denote larger of the two no.
X can be 2, 3, 4 or 5.
P(X = 2) = P(larger no. is 2) = {(1, 2) and (2, 1)}
$=\frac{2}{30}$
$\text{P}(\text{X}=3)=\frac{4}{30}$
$\text{P}(\text{X}=4)=\frac{6}{30}$
$\text{P}(\text{X}=5)=\frac{8}{30}$
$\text{Mean}=\text{E}(\text{X})=2\times\frac{2}{30}+3\times\frac{4}{30}+4\times\frac{6}{30}+5\times\frac{8}{3 0}$
$=\frac{4+12+24+40}{30}$
$=\frac{80}{30}$
$\text{Variance}=2^2\times\frac{2}{30}+3^2\times\frac{4}{30}+4^2\times\frac{6}{30}+5^2\times\frac{8}{30}$
$=\frac{8+36+96+200}{30}$
$=\frac{340}{30}=\frac{34}{3}$
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Question 575 Marks
Suppose a girl throws a die. If she gets $1$ or $2,$ she tosses a coin three times and notes the number of tails. If she gets $3, 4, 5$ or $6,$ she tosses a coin once and notes whether a ‘head’ or ‘tail’ is obtained. If she obtained exactly one ‘tail’, what is the probability that she threw $3, 4, 5$ or $6$ with the die?
Answer
Let $E_1$ be the event that the girl. Gets $1$ or $2$ on the roll
$\text{P(E}_1)=\frac{2}{6}=\frac{1}{3}$
Let $E_2$ be the event that the girl gets $3, 4, 5$ or $6$ on the roll $\text{P(E}_2)=\frac{4}{6}=\frac{2}{3}$
Let $A$ be event that she obtained exactly one tails
If she tossed a coin $3$ times & exactly $1$ tail shows then $[HTH, HHT, THH] = 3$
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{3}{8}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{1}{2} ($If she tossed a coin only once & exactly $1$ shows$)$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{1}{2}\times\frac{2}{3}}{\frac{1}{2}\times\frac{2}{3}+\frac{3}{8}\times\frac{1}{3}}=\frac{8}{11}$
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Question 585 Marks
A manufacturer has three machine operators $A, B$ and $C.$ The first operator $A$ produces $1\%$ of defective items, whereas the other two operators $B$ and $C$ produces $5\%$ and $7\%$ defective items respectively. $A$ is on the job for $50\%$ of the time, $B$ on the job $30\%$ of the time and $C$ on the job for $20\%$ of the time. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by $A$?
Answer
Let $E_1, E_2$ and $E_3$ be the event that machine is operated by $A, B,$ and $C$ respectively.
Let $A$ be the event of producing defective items.
$\therefore\text{P}(\text{E}_1)=50\%=\frac{1}{2}$
$\text{P}(\text{E}_2)=30\%=\frac{3}{10}$
$\text{P}(\text{E}_3)=20\%=\frac{1}{5}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=1\%=\frac{1}{100}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=5\%=\frac{5}{100}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=7\%=\frac{7}{100}$
Using Bayes' theorem, we get
Required probability $=\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\big(\frac{\text{A}}{\text{E}_1}\big)}{\text{P}(\text{E}_1)\text{P}\big(\frac{\text{A}}{\text{E}_1}\big)+\text{P}(\text{E}_2)\text{P}\big(\frac{\text{A}}{\text{E}_2}\big)+\text{P}(\text{E}_3)\text{P}\big(\frac{\text{A}}{\text{E}_3}\big)}$
$=\frac{\frac{1}{2}\times\frac{1}{100}}{\frac{1}{2}\times\frac{1}{100}+\frac{3}{10}\times\frac{5}{100}+\frac{1}{5}\times\frac{7}{100}}$
$=\frac{5}{34}$
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Question 595 Marks
Two cards are selected at random from a box which contains five cards numbered $1, 1, 2, 2,$ and $3.$ Let $X$ denote the sum and $Y$ the maximum of the two numbers drawn. Find the probability distribution, mean and variance of $X$ and $Y.$
Answer
Box contains five cards $1, 1, 2, 2, 3$.
Here, $X$ denotes the sum of the two number on cards drawn.
$Y$ denotes the maximum of the two number.
So, $X = 2, 3, 4, 5 Y = 1, 2, 3 P(X = 2) = P(1)P(1) =\frac{2}{5}\times\frac{1}{4}$
$=0.1 P(X = 3) = P(1)P(2) + P(2)P(1) =\frac{2}{5}\times\frac{2}{4}+\frac{2}{5}\times\frac{2}{4}$
$=0.4 P(X = 4) = P(2)P(2) + P(1)P(3) + P(3)P(1) =\frac{2}{5}\times\frac{1}{4}+\frac{2}{5}\times\frac{1}{4}+\frac{1}{5}\times\frac{2}{4}$
$=0.3 P(X = 5) = P(2)P(3) + P(3)P(2) =\frac{2}{5}\times\frac{2}{4}+\frac{1}{5}\times\frac{2}{4}$
$=0.2 Probability distribution for X$
$x:$ $2$ $3$ $4$ $5$
$P(x):$ $0.1$ $0.4$ $0.3$ $0.2$
Now,
$x_i$ $p_i$ $x_ip_i$ $x_i^2p_i$
$2$ $0.1$ $0.1$ $0.4$
$3$ $0.4$ $1.2$ $3.6$
$4$ $0.3$ $1.2$ $4.8$
$5$ $0.2$ $1.0$ $5.0$
$
$		 $
$		 $\sum\text{xp}=3.6$ $\sum\text{x}^2\text{p}=13.8$
Mean $=\sum\text{xp}$ Mean $= 3.6$ Variance $=\sum\text{x}^2\text{p}-(\text{mean})^2 =13.8-(3.6)^2 =13.8-12.96$ Variance $= 0.84 P(Y = 1) = P(1)P(1) =\frac{2}{5}\times\frac{1}{4} =\frac{2}{20} =0.1 P(Y = 2) = P(1)P(2) + P(2)P(1) + P(2)P(2)$
$ =\frac{2}{5}\times\frac{2}{4}+\frac{2}{5}\times\frac{2}{4}+\frac{2}{5}\times\frac{1}{4}$
$=0.5 P(Y = 3) = P(1)P(3) + P(2)P(3) + P(3)P(1) + P(3)P(2)$
$=\frac{2}{5}\times\frac{1}{4}+\frac{2}{5}\times\frac{1}{4}+\frac{1}{5}\times\frac{2}{4}+\frac{1}{5}\times\frac{2}{4}$
 $=0.4$ Probability distribution for $Y$ is
$x:$ $1$ $2$ $3$
$p(x):$ $0.1$ $0.5$ $0.4$
 
$y_i$ $p_i$ $y_ip_i$ $y_i^2p_i$
$1$ $0.1$ $0.1$ $0.1$
$2$ $0.5$ $1.0$ $2.0$
$3$ $0.4$ $1.2$ $3.6$
$
$		 $
$		 $\sum\text{xp}=2.3$ $\sum\text{x}^2\text{p}=5.7$
Mean $=\sum\text{xp}=2.3$ Variance $=\sum\text{x}^2\text{p}-(\text{mean})^2 =5.1-(2.3)^2$ Variance $= 0.41$
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Question 605 Marks
In a game, a man wins Rs 5 for getting a number greater than 4 and loses Rs 1 otherwise, when a fair die is thrown. The man decided to thrown a die thrice but to quit as and when he gets a number greater then 4. Find the expected value of the amount he wins or loses.
Answer
The man may get number greater than 4 in the first throw and then he quits the game. He may get a number less than equat to 4. In the first throw and in the second throw he may get the number greater than 4 and quits the game.
In the first two throws he gets a number less than equal to 4 and in the third throw he may get a number greater than 6. He may not get number greater than 4 in any one of three throws.
Let X be the amount he wins/looses.
Then, X can take values -3, 3, 4, 5 such that
P(X= 5) = P(Getting number greater than 4 in first throw) $=\frac{1}{3}$
(X= 4) = P(Getting number less than equal to 4 in the first throw and number greater than 4 in second throw) $=\frac{4}{6}\times\frac{2}{6}=\frac{2}{9}$
P (X = 3) = P(Getting number less than equal to 4 in the first two throws and number greater than 4 in third throw) $=\frac{4}{6}\times\frac{4}{6}\times\frac{2}{6}=\frac{4}{27}$
P(X = -3) = P(Getting number less than equal to 4 in all three throws) $=\frac{4}{6}\times\frac{4}{6}\times\frac{4}{6}=\frac{8}{27}$
$\text{X}$ $5$ $4$ $3$ $-3$
$\text{P}(\text{X})$ $\frac{1}{3}$ $\frac{2}{9}$ $\frac{4}{27}$ $\frac{8}{27}$
$\text{E}(\text{X})=\Big(5\times\frac{1}{3}\Big)+4\Big(\frac{2}{9}\Big)+3\Big(\frac{4}{27}\Big)-3\Big(\frac{8}{27}\Big)$
$=\frac{1}{27}(45+24+12-24)$
$=\frac{57}{27}$
$=\frac{19}{9}$
Expected value of the amount he wins or loses is $=\frac{19}{9}$
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Question 615 Marks
Prove that:
  1. $\text{P}(\text{A})=\text{P}(\text{A}\cap\text{B})+\text{P}(\text{A}\cap\bar{\text{B}})$
  2. $\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A}\cap\text{B})+\text{P}(\text{A}\cap\bar{\text{B}})+\text{P}(\bar{\text{A}}\cap\text{B})$
Answer
  1. $\because\text{P}(\text{A})=\text{P}(\text{A}\cap\text{B})+\text{P}(\text{A}\cap\bar{\text{B}})$
$\therefore\text{R.H.S.}=\text{P}(\text{A}\cap\text{B})+\text{P}(\text{A}\cap\bar{\text{B}})$
$=\text{P}(\text{A})\cdot\text{P}(\text{B})+\text{P}(\text{A})\cdot\text{P}\bar{(\text{B})}$
$=\text{P}(\text{A})\big[\text{P}(\text{B})+\text{P}\bar{(\text{B})}\big]$
$=\text{P}(\text{A})\big[\text{P}(\text{B})+1-\text{P}(\text{B})\big]$ $\big[\because\text{P}\bar{(\text{B})}=1-\text{P}(\text{B})\big]$
$= \text{P(A) = L. H. S} $ Hence proved.
  1. $\because\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A}\cap\text{B})+\text{P}(\text{A}\cap\bar{\text{B}})+\text{P}(\bar{\text{A}}\cap\text{B})$
$\therefore\text{R.H.S.}=\text{P}(\text{A})\cdot\text{P}(\text{B})+\text{P}(\text{A})\cdot\text{P}\bar{(\text{B})}+\text{P}\bar{(\text{A})}\cdot\text{P}(\text{B})$
$=\text{P}(\text{A})\cdot\text{P}(\text{B})+\text{P}(\text{A})\cdot\big[1-\text{P}(\text{B})\big]+\big[1-\text{P}(\text{A})\big]\text{P}(\text{B})$
$=\text{P}(\text{A})\cdot\text{P}(\text{B})+\text{P}(\text{A})-\text{P}(\text{A})\cdot\text{P}(\text{B})+\text{P}(\text{B})-\text{P}(\text{A})\cdot\text{P}(\text{B})$
$=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A})\cdot\text{P}(\text{B})$
$=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$=\text{P}(\text{A}\cup\text{B})=\text{L.H.S.}$ Hence proved.
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Question 625 Marks
Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as
  1. number greater than 4
  2. six appears on at least one die
Answer

S = {1, 2, 3, 4, 5, 6} ⇒ n(S) = 6
  1. Let A be the set of favourable events. ⇒ n(A) = 2
$\therefore\ \text{P}(\text{A})=\frac{\text{n}(\text{S})}{\text{n}(\text{A})}=\frac{2}{6}=\frac{1}{3}\ \text{and}\ \text{P}(\overline{\text{A}})=1-\frac{1}{3}=\frac{2}{3}$
n = 2, r = 0, 1, 2
$\text{P}(\text{X}=0)=\text{P}(\overline{\text{A}}).\text{P}(\overline{\text{A}})=\frac{2}{3}\times\frac{2}{3}=\frac{4}{9}$
$\text{P}(\text{X}=1)=2\text{P}(\text{A}).\text{P}(\overline{\text{A}})=2\times\frac{1}{3}\times\frac{2}{3}=\frac{4}{9}$
$\text{P}(\text{X}=2)=\text{P}(\text{A}).\text{P}(\text{A})=\frac{1}{3}\times\frac{1}{3}=\frac{1}{9}$
Probability distribution
$\text{X}$ $0$ $1$ $2$
$\text{P}(\text{X})$ $\frac{4}{9}$ $\frac{4}{9}$ $\frac{1}{9}$
  1. Let A represents that 6 appears on one die ⇒ A = {6} ⇒ n(A) = 1
$\therefore\ \text{P}(\text{A})=\frac{\text{n}(\text{S})}{\text{n}(\text{A})}=\frac{1}{6}\ \text{and}\ \text{P}(\overline{\text{A}})=1-\frac{1}{6}=\frac{5}{6}$
n = 2, r = 0, 1, 2
$\text{P}(\text{X}=0)=\text{P}(\overline{\text{A}}).\text{P}(\overline{\text{A}})=\frac{5}{6}\times\frac{5}{6}=\frac{25}{36}$
$\text{P}(\text{X}=1)=2\text{P}(\text{A}).\text{P}(\overline{\text{A}})=2\times\frac{1}{6}\times\frac{5}{6}=\frac{10}{36}$
$\text{P}(\text{X}=2)=\text{P}(\text{A}).\text{P}(\text{A})=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$
P(at least on six) $=\frac{10}{36}\times\frac{1}{36}=\frac{11}{36}$
Probability distribution
$\text{X}_i$ $0$ $1$
$\text{P}_i$ $\frac{25}{36}$ $\frac{11}{36}$
 
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Question 635 Marks
Find the mean variance and standard deviation of the following probability distribution
$x_i$ $a$ $b$
$p_i$ $p$ $q$
Where $p + q = 1$
Answer
$x_i$ $p_i$ $p_ix_i$ $p_ix_i^2$
$a$ $p$ $ap$ $a^2p$
$b$ $q$ $bq$ $b^2q$
    $\sum\text{p}_\text{i}\text{x}_\text{i}=\text{ap}+\text{bq}$ $\sum\text{p}_\text{i}\text{x}_\text{i}^2=\text{a}^2\text{p}+\text{b}^2\text{q}$
Now,
Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=\text{ap}+\text{bq}$
Variance $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-(\text{Mean})^2$
$= a^2p + b^2q - (ap + bq)^2$
$= a^2p + b^2q - a^2p^2 - b^2q^2 - 2abpq$
$= a^2p - a^2p^2 + b^2q - b^2q^2 - 2abpq$
$= a^2p(1 - p) + b^2q(1 - q) - 2abpq$
$= a^2pq + b^2qp - 2abpq (\because\ \text{p}+\text{q}=1)$
$= pq(a^2 + b^2 - 2ab)$
$= pq(a - b)^2$ 
Step Deviation $=\sqrt{\text{Variance}}$
$=\sqrt{\text{pq}(\text{a-b})^2}$
$=|\text{a}-\text{b}|\sqrt{\text{pq}}$
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Question 645 Marks
Two numbers are selected at random from integers 1 through 9. If the sum is even, find the probability that both the numbers are odd.
Answer
Two numbers are selected at random from integers 1 through 9.
A = Both numbers are odd
A = {(3, 1), (5, 1), (7, 1), (9, 1), (3, 5), (3, 7), (9, 3), (5, 3), (5, 7), (5, 9), (7, 3), (7, 5), (7, 9), (9, 3), (9, 5), (9, 7)}
B = Sum of both numbers is even
A = Sum of both numbers is 2, 4, 6, 8, 10, 12, 14, 16 or 18 = {(1, 3), (1, 5), (2, 4), (1, 7), (2, 6), (3, 5), (1, 9), (2, 8), (3, 7), (4, 6), (7, 5), (8, 4), (9, 3), (8, 6), (9, 5), (9, 7)}
$(\text{A}\cap\text{B})=$ {(1, 3), (1, 5), (1, 7), (3, 5), (1, 9), (3, 7), (7, 5), (9, 3), (9, 5), (9, 7)} 
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}$
$=\frac{10}{16}$
Required probability $=\frac{10}{16}$
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Question 655 Marks
A fair coin and an unbiased die are tossed. Let $A$ be the event ‘head appears on the coin’ and $B$ be the event ‘$3$ on the die’. Check whether $A$ and $B$ are independent events or not.
Answer
If a fair coin and an unbiased die are tossed, then the sample space $S$ is given by, $\text{S} = \{(\text{H, 1}),\ (\text{H, 2}),\ (\text{H, 3}),\ (\text{H, 4}),\ (\text{H, 5}),\ (\text{H, 6}),\text{(T, 1}),\ \text{(T, 2}),\ \text{(T, 3}),\ \text{(T, 4}),\ (\text{T, 5}),\ \text{(T, 6})\}$
Let $A$ : Head appears on the coin
$\text{A}=\left\{(\text{H, 1}),\ (\text{H, 2}),\ (\text{H, 3}),\ (\text{H, 4}),\ (\text{H, 5}), (\text{H, 6})\right\}$
$\Rightarrow\text{P}(\text{A})=\frac{6}{12}=\frac{1}{2}$
$B: 3$ on die $=\left\{(\text{H, 3}),\ (\text{T, 3})\right\}$
$ \text{P}(\text{B})=\frac{2}{12}=\frac{1}{6}$
$\therefore\ \text{A}\cap\text{B}=\left\{(\text{H, 3})\right\}$
$\text{P}(\text{A}\cap\text{B})=\frac{1}{12}$
$\text{P}(\text{A}).\text{P}(\text{B})=\frac{1}{2}\times\frac{1}{6}=\text{P}(\text{A}\cap\text{B})$
Therefore,$ A$ and $B$ are independent events.
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Question 665 Marks
An insurance company insured $2000$ scooter drivers, $4000$ car drivers and 6000 truck drivers. The probability of an accidents are $0.01, 0.03$ and $0.15$ respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Answer
Let $E_{1 }=$ Person chosen is a scooter driver,
$E_{2 }=$ Person chosen is a car driver,
$E_{3 }=$ Person chosen is a truck driver and
$A =$ Person meets with an accidentSince there are $12000$ persons, therefore,
$\text{Now}\ \ \ \text{P}(\text{E}_1)=\frac{2000}{12000}=\frac{1}{6},\ \text{P}(\text{E}_2)=\frac{4000}{12000}=\frac{1}{3},\ \text{P}(\text{E}_3)=\frac{6000}{12000}=\frac{1}{2}$
It is given that $\text{P}(\text{A}|\text{E}_1) = P($a person meets with an accident, he is a scooter driver$) = 0.01$
Similarly, $\text{P}(\text{A}|\text{E}_2)=0.03\ \text{and}\ \text{P}(\text{A}|\text{E}_3)=0.15$
To find: $P($ person meets with an accident that he was a scooter driver$)$
Therefore, by Bayes’ theorem,
$\text{P}(\text{E}_1|\text{A})=\frac{\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)}{\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\text{P}(\text{A}|\text{E}_2)+\text{P}(\text{E}_3)(\text{A}|\text{E}_3)}$
$=\frac{\frac{1}{6}\times0.01}{\frac{1}{6}\times0.01+\frac{1}{3}\times0.03+\frac{1}{2}\times0.15}=\frac{1}{1+6+45}=\frac{1}{52}$
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Question 675 Marks
In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English news papers. A student is selected at random.
  1. Find the probability that she reads neither Hindi nor English news papers.
  2. If she reads Hindi news paper, find the probability that she reads English news paper.
  3. If she reads English news paper, find the probability that she reads Hindi news paper.
Answer
Let H denote the students who read Hindi newspaper and E denote the students who read English newspaper.It is given that,
$\text{P}(\text{H})=60\%=\frac{60}{100}=\frac{6}{10}=\frac{3}{5}$
$\text{P}(\text{E})=40\%=\frac{40}{100}=\frac{2}{5}$
$\text{P}(\text{H}\cap\text{E})=20\%=\frac{20}{100}=\frac{1}{5}$
  1. Probability that a student reads Hindi or English newspaper is,
$\text{P}(\text{H}\cup\text{E})'=1-\text{P}(\text{H}\cup\text{E})$
$=1-\big\{\text{P}(\text{H})+\text{P}(\text{E})-\text{P}(\text{H}\cap\text{E})\big\}$
$=1-\Big(\frac{3}{5}+\frac{2}{5}-\frac{1}{5}\Big)$
$=1-\frac{4}{5}$
$=\frac{1}{5}$
  1. Probability that a randomly chosen student reads English newspaper, if she reads Hindi news paper, is given by P(E|H).
$\text{P}(\text{E}|\text{H})=\frac{\text{P}(\text{E}\cap\text{H})}{\text{P}(\text{H})}$
$=\frac{\frac{1}{5}}{\frac{3}{5}}$
$=\frac{1}{3}$
  1. Probability that a randomly chosen student reads Hindi newspaper, if she reads English newspaper, is given by P(H|E).
$\text{P}(\text{H}|\text{E})=\frac{\text{P}(\text{H}\cap\text{E})}{\text{P}(\text{E})}$
$=\frac{\frac{1}{5}}{\frac{2}{5}}$
$=\frac{1}{2}$
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Question 685 Marks
A random variable $X$ has the following probability distribution:
$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$
Determine
  1. $k$
  2. $P(X < 3)$
  3. $P(X > 6)$
  4. $P(0 < X < 3)$
Answer
  1. Since, the sum of all the probabilities of a distribution is $1.$
$\therefore P(X = 0) + P(X = 1) + …. + P(X = 7) = 1$
$\Rightarrow 0 + k + 2k + 2k + 3k + k^2 + 2k^{2 }+ 7k^{2 }+ k = 1$
$\Rightarrow 10k^2 + 9k - 1 = 0$
$\Rightarrow (10k - 1) (k + 1) = 0$
$\Rightarrow 10k - 1 = 0$ or $k + 1 = 0$
$\Rightarrow\ \text{k}=\frac{1}{10}$ or $k = - 1$
Since, $k \geq 0,$ therefore $k = − 1$ is not possible.
$\therefore\ \text{k}=\frac{1}{10}$
  1. $P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$
$= 0 + k + 2k$
$=3\text{k}=3\times\frac{1}{10}=\frac{3}{10}$
  1. $P(X > 6) = P(X = 7)$
$=7\text{k}^2+\text{k}=7\Big(\frac{1}{10}\Big)^2+\frac{1}{10}=\frac{17}{100}$
  1. $P(0 < X < 3) = P(X = 1) + P(X = 2)$
$= k + 2k = 3k = \frac{3}{10}$
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Question 695 Marks
Two numbers are selected at random (without replacement) from positive integers 2, 3, 4, 5, 6 and 7. Let X denote the larger of the two numbers obtained. Find the mean and variance of the probability distribution of X.
Answer
We can select two positive in 6 × 5 = 30 different waysP(X = 3) = P(larger number is 3) = {(2, 3), (3, 2)} $=\frac{2}{30}$
P(X = 4) = P(larger number is 4) = {(2, 4), (4, 2), (3, 4), (4, 3)} $=\frac{4}{30}$
P(X = 5) = P(larger number is 5) = {(2, 5), (5, 2), (3, 5), (5, 3), (4, 5), (5, 4)} $=\frac{6}{30}$
P(X = 6) = P(larger number is 6) = {(2, 6), (6, 2), (3, 6), (6, 3), (4, 6), (6, 4), (5, 6), (6, 5)} $=\frac{8}{30}$
P(X = 7) = P(larger number is 7) = {(2, 7), (7, 2), (3, 7), (7, 3), (4, 7), (7, 4), (5, 7), (7, 5), (6, 7), (7, 6)} $=\frac{10}{30}$
Thus, the probability distribution of random variable X is,
$\text{x}_\text{i}$ $\text{p}_\text{i}$ $\text{x}_\text{i}\text{p}_\text{i}$ $\text{x}_\text{i}^2\text{p}_\text{i}$
$3$ $\frac{2}{30}$ $\frac{6}{30}$ $\frac{18}{30}$
$4$ $\frac{4}{30}$ $\frac{16}{30}$ $\frac{64}{30}$
$5$ $\frac{6}{30}$ $\frac{30}{30}$ $\frac{150}{30}$
$6$ $\frac{8}{30}$ $\frac{48}{30}$ $\frac{288}{30}$
$7$ $\frac{10}{30}$ $\frac{70}{30}$ $\frac{490}{30}$
    $\sum\text{x}_\text{i}\text{p}_\text{i}=\frac{17}{3}$ $\sum\text{x}_\text{i}\text{p}_\text{i}^2=\frac{101}{3}$
Mean $=\sum\text{x}_\text{i}\text{p}_\text{i}=\frac{17}{3}$
Variance $=\sum\text{x}_\text{i}\text{p}_\text{i}-\big(\sum\text{x}_\text{i}\text{p}_\text{i}\big)^2=\frac{101}{3}-\Big(\frac{17}{3}\Big)=\frac{14}{9}$
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Question 705 Marks
Assume that each born child is equally likely to be a boy or a girl. If a family has two children, then what is the constitutional probability that both are girls? Given that.
  1. The youngest is a girl,
  2. At least one is a girl.
Answer
  1. Let 'A' be the event that both the children born are girls. Let 'B' be the event that the youngest is a girls. We have to find conditional probability $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$.
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$
$\text{A}\subset\text{B} \Rightarrow\text{A}\cap\text{B}=\text{A}$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\text{P(A)}=\text{P(GG)}=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$
$\text{P(B)}=\text{P(BG)}+\text{P(GG)}=\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2} \\=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$
Hence, $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}$
  1. Let 'A' be the event that both the children born are girls. Let 'B' be the event that at least one is a girl. We have to find the conditional probability $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$.
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\text{A}\subset\text{B}\Rightarrow\text{A}\cap\text{B}=\text{A}$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\text{P(A)}=\text{P(GG)}=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$
$\text{P(B)}=1-\text{P(BB)}=1-\frac{1}{2}\times\frac{1}{2}=1-\frac{1}{4}=\frac{3}{4}$
Hence, $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$
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Question 715 Marks
A pair of dice is thrown. Find the probability of getting the sum 8 or more, if 4 appears on the first die.
Answer
A pair of dice is thrownA = getting sum 8 or more
= Getting sum 8, 9, 10, 11 or 12 on the pair of dice
A = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (3, 6)
(4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4)
(5, 6), (6, 5), (6, 6)
B = 4 on first die
B = {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)}
$(\text{A}\cap\text{B})=\{(4, 4), (4, 5), (4, 6)\}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}$
$=\frac{3}{6}$
$=\frac{1}{2}$
Required probability $=\frac{1}{2}$
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Question 725 Marks
Three cards are drawn successively, without replacement from a pack of 52 well shuffled cards. What is the probability that first two cards are kings and third card drawn is an ace?
Answer
Let K denote the event that the card drawn is king and a be the event taht the card drawn is an ace.
We are ot find P (K K A).
Now, $\text{P(K)}=\frac{4}{52}$
Also, $\text{P}\Big(\frac{\text{K}}{\text{K}}\Big)$ is the probability of second king with the condition that one king has already been drawn.
Now, there are 3 king in (52 - 1) = 51 cards.
$\therefore\ \text{P} \Big(\frac{\text{K}}{\text{K}}\Big)=\frac{3}{51}$
Lastly, $\text{P}\Big(\frac{\text{A}}{\text{K K}}\Big)$ is the probability of third drawn card to be an ace, woth the condition that two kings have already been drawn.
Now, there are four aces in left 50 cards.
$\therefore\ \text{P}\Big(\frac{\text{A}}{\text{K K}}\Big)=\frac{4}{50}$
By multiplication law of probability, we have
$\text{P(K K A)}=\text{P(K) P}\Big(\frac{\text{K}}{\text{K}}\Big) \text{ P}\Big(\frac{\text{A}}{\text{KK}}\Big)$
$=\frac{4}{52}\times\frac{3}{51}\times\frac{4}{50}=\frac{2}{5525}$
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Question 735 Marks
An anti-aircraft gun can take a maximum of 4 shots at an enemy plane moving away from it. The probabilities of hitting the plane at the first, second, third and fourth shot are 0.4, 0.3, 0.2 and 0.1 respectively. What is the probability that the gun hits the plane?
Answer
Given,
An anti aircraft gun can rake a maximum 4 shots at an enemy plane
Consider,
A = Htting the plane at first shot
B = Hetting the plane at second shot
C = Hetting the place at third shot
D = Hetting the place at fourth shot
⇒ P(A) = 0.4, P(B) = 0.3, P(C) = 0.2, P(D) = 0.1
P (Gun hits the place)
= 1 - P(Gun does not hit the plane)
= 1 - P(Non of the foru shots hot the place)
$=1-\text{P}(\overline{\text{A}}\cap\overline{\text{B}}\cap\overline{\text{C}}\cap\overline{\text{D}})$
$=1-\text{P}(\overline{\text{A}})\text{ P}(\overline{\text{B}})\text{ P}(\overline{\text{C}})\text{ P}(\overline{\text{D}})$
$=1-[1-\text{P(A)}]\big[1-\text{P}(\overline{\text{B}})\big][1-\text{P(C)}][1-\text{P(D)}]$
$=1-[1-0.4][1-0.3][1-0.2][1-0.1]$
$=1-(0.6)(0.7)(0.8)(0.9)$
$=1.03024$
$=0.6976$
Required probability = 0.6976
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Question 745 Marks
Arun and Tarun appeared for an interview for two vacancies. The probability of Arun's selection is $\frac{1}{4}$ and that to Tarun's rejection is $\frac{2}{3}$. Find the probability that at least one of them will be selected.
Answer
Given,
Probability of Arun's (A) selection $=\frac{1}{4}$
$\text{P(A)}=\frac{1}{4}$
Probability of tarun's (T) rejection $=\frac{2}{3}$
$\text{P}(\overline{\text{T}})=\frac{2}{3}$
$\text{P}(\overline{\text{A}})=1-\text{P(A)}$
$\Rightarrow\ \text{P}(\overline{\text{A}})=1-\frac{1}{4}$
$\Rightarrow\ \text{P}(\overline{\text{A}})=\frac{3}{4}$
$\text{P(T)}=1-\text{P}(\overline{\text{T}})$
$\Rightarrow\ \text{P(T)}=1-\frac{2}{3}$
$\Rightarrow\ \text{P(T)}=\frac{1}{3}$
P (At least one of them will be selelcted)
= 1 - P(None of them selected)
$=1-\text{P}(\overline{\text{A}}\cap\overline{\text{T}})$
$=1-\text{P}(\overline{\text{A}})\text{ P}(\overline{\text{T}})$
$=1-\frac{2}{3}\times\frac{3}{4}$
$=\frac{1}{2}$
Required probability $=\frac{1}{2}$
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Question 755 Marks
One bag contains $4$ white and $5$ black balls. Another bag contains $6$ white and $7$ black balls. A ball is transferred from first bag to the second bag and then a ball is drawn from the second bag. Find the probability that the ball drawn is white.
Answer
There are two bags.
Bag $(1)$ contain $4$ white and $5$ black balls.
Bag $(2)$ contain $6$ white and $7$ black balls.
A ball is taken from bag $(1)$ and without seeing its colour is pur in bag $(2).$
Then a ball is drawn from bag $(2)$ and is found white in colour.
$P(1$ white ball from bag $1) =\frac{4}{9}$
$\text{P}(\text{W}_1)=\frac{4}{9}$
$P(1$ black ball from bag $1) =\frac{5}{9}$
$\text{P}(\text{B}_1)=\frac{5}{9}$
$P(1$ white ball from bag $2$ given $W_1$ is put in bag $2)$
$\text{P}\Big(\frac{\text{W}_2}{\text{W}_1}\Big)=\frac{7}{14}$
$\text{P}\Big(\frac{\text{W}_2}{\text{W}_1}\Big)=\frac{1}{2}$
$P(1$ white ball from bag $2$ given $B_1$ is put in bag $2)$
$\text{P}\Big(\frac{\text{W}_2}{\text{B}_1}\Big)=\frac{6}{14}$
$P(1$ white from bag $2)$
$=\text{P}(\text{W}_1)\text{P}\Big(\frac{\text{W}_2}{\text{W}_1}\Big)+\text{P}(\text{B}_1)\text{P}\Big(\frac{\text{W}_2}{\text{W}_1}\Big)$
$=\frac{4}{9}\times\frac{1}{2}+\frac{5}{9}\times\frac{6}{14}$
$=\frac{4}{18}+\frac{30}{126}$
$=\frac{58}{126}$
$=\frac{29}{63}$
Required probability $=\frac{29}{63}$
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Question 765 Marks
Probability of solving specific problem independently by A and B are $\frac{1}{2}\ \text{and}\ \frac{1}{3}$ respectively.If both try to solve the problem independently, find the probability that
  1. The problem is solved.
  2. Exactly one of them solves the problem.
Answer
Probability of solving the problem by A, P(A) $=\frac{1}{2}$Probability of solving the problem by B, P(B) $=\frac{1}{3}$
Since the problem is solved independently by A and B,
$\therefore\text{P}(\text{AB})=\text{P}(\text{A})\cdot\text{P}(\text{B})=\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}$
$\text{P}(\text{A}')=1-\text{P}(\text{A})=1-\frac{1}{2}=\frac{1}{2}$
$\text{P}(\text{B}')=1-\text{P}(\text{B})=1-\frac{1}{3}=\frac{2}{3}$
  1. Probability that the problem is solved $=\text{P}(\text{A}\cup\text{B})$
$=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{AB})$
$=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}$
$=\frac{4}{6}$
$=\frac{2}{3}$
  1. Probability that exactly one of them solves the problem is given by,
$\text{P}(\text{A}).\text{P}(\text{B}')+\text{P}(\text{B}).\text{P}(\text{A}')$
$ =\frac{1}{2}\times\frac{2}{3}+\frac{1}{2}\times\frac{1}{3}$
$=\frac{1}{3}+\frac{1}{6}$
$=\frac{1}{2}$
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Question 775 Marks
A class has 15 students whose ages are 14, 17, 15, 14, 21, 19, 20, 16, 18, 17, 20, 17, 16, 19 and 20 years respectively. One student is selected in such a manner that each has the same chance to being selected and the age X of the selected student is recorded. What is the probability distribution of the random variable X.
Answer
Here, X denote the number of two number or two dice thrown together.
So, X = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
So,
$\text{P}(\text{X}=2)=\frac{1}{36}$ [Possible pairs: (1, 1)]
$\text{P}(\text{X}=3)=\frac{2}{36}=\frac{1}{18}$ [Possible pairs: (1, 2), (2,1)]
$\text{P}(\text{X}=4)=\frac{3}{36}=\frac{1}{12}$ [Possible pairs: (1, 3), (2,2), (3, 1)]
$\text{P}(\text{X}=5)=\frac{4}{36}=\frac{1}{9}$ [Possible pairs: (1, 4), (2, 3), (3, 2), (4, 1)]
$\text{P}(\text{X}=6)=\frac{5}{36}$ [Possible pairs: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1)]
$\text{P}(\text{X}=7)=\frac{6}{36}=\frac{1}{6}$ [Possible pairs: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)]
$\text{P}(\text{X}=8)=\frac{5}{36}$ [Possible pairs: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)]
$\text{P}(\text{X}=9)=\frac{4}{36}=\frac{1}{9}$ [Possible pairs: (3, 6), (4, 5), (5, 4), (6, 3)]
$\text{P}(\text{X}=10)=\frac{3}{36}=\frac{1}{12}$ [Possible pairs: (4, 6), (5, 5), (6, 4)]
$\text{P}(\text{X}=11)=\frac{2}{36}=\frac{1}{18}$ [Possible pairs: (5, 6), (6,5)]
$\text{P}(\text{X}=12)=\frac{1}{36}$ [Possible pairs: (6, 6)]
So, required possibility distribution is
$\text{X}:$
$2$
$3$
$4$
$5$
$6$
$7$
$8$
$9$
$10$
$11$
$12$
$\text{P}(\text{X}):$
$\frac{1}{36}$
$\frac{1}{18}$
$\frac{1}{12}$
$\frac{1}{9}$
$\frac{5}{36}$
$\frac{1}{6}$
$\frac{5}{36}$
$\frac{1}{9}$
$\frac{1}{12}$
$\frac{1}{18}$
$\frac{1}{36}$
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Question 785 Marks
Find the probability that the sum of the numbers showing on two dice is 8, given that at least one die does not show five.
Answer
Two dice are thrown
A = Sum of the numbers on dice is 8
A = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
B = At least one die does not show five
B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 6)}
$(\text{A}\cap\text{B})=\{(2, 6), (4, 6), (6, 2)\}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}$
$=\frac{3}{25}$
Require probability $=\frac{3}{25}$
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Question 795 Marks
A shopkeeper sells three types of flower seeds $A_1, A_2$ and $A_3.$ They are sold as a mixture where the proportions are $4 : 4 : 2$ respectively. The germination rates of the three types of seeds are $45\%, 60\%$ and $35\%.$ Calculate the probability:
  1. Of a randomly chosen seed to germinate.
  2. That it will not germinate given that the seed is of type $A_{3.}$
  3. That it is of the type $A_2$ given that a randomly chosen seed does not germinate.
Answer
We have, $\text{P}(\text{A}_1)=\frac{4}{10},\text{P}(\text{A}_2)=\frac{4}{10}$ and $\text{P}(\text{A}_3)=\frac{2}{10}$
Where $A_1, A_{2 }$ and $A_{3 }$ denote the three types of flower seeds.
Let $E$ be the event that a seed germinates and be the event that a seed does not germinate.
$\therefore\text{P}\Big(\frac{\text{E}}{\text{A}_1}\Big)=\frac{45}{100},$
$\text{P}\Big(\frac{\text{E}}{\text{A}_2}\Big)=\frac{60}{100}$ 
and $\text{P}\Big(\frac{\text{E}}{\text{A}_3}\Big)=\frac{35}{100}$
and $\text{P}\Big(\frac{\text{E}'}{\text{A}_1}\Big)=\frac{55}{100},\text{P}\Big(\frac{\text{E}'}{\text{A}_2}\Big)=\frac{40}{100}$ 
and $\text{P}\Big(\frac{\text{E}'}{\text{A}_3}\Big)=\frac{65}{100}$
  1. $\therefore\text{P}(\text{E})=\text{P}(\text{A}_1)\cdot \text{P}\Big(\frac{\text{E}}{\text{A}_1}\Big)+\text{P}(\text{A}_2)\cdot\text{P}\Big(\frac{\text{E}}{\text{A}_2}\Big) +\text{P}(\text{A}_3)\cdot\text{P}\Big(\frac{\text{E}}{\text{A}_3}\Big)$
$=\frac{4}{10}\cdot\frac{45}{100}+\frac{4}{10}\cdot\frac{60}{100}+\frac{2}{10}\cdot\frac{35}{100}$
$=\frac{180}{1000}+\frac{240}{1000}+\frac{70}{1000}$
$=\frac{490}{1000}=0.49$
  1. $\text{P}\Big(\frac{\text{E}'}{\text{A}_3}\Big)=1-\text{P}\Big(\frac{\text{E}}{\text{A}_3}\Big)$
$=1-\frac{35}{100}=\frac{65}{100}$
  1. $\text{P}\Big(\frac{\text{A}_2}{\text{E}'}\Big)=\frac{\text{P}(\text{A}_2)\cdot\text{P}\Big(\frac{\text{E}'}{\text{A}_2}\Big)}{\text{P}(\text{A}_1)\cdot\text{P}\Big(\frac{\text{E}'}{\text{A}_1}\Big)+\text{P}(\text{A}_2)\cdot\text{P}\Big(\frac{\text{E}'}{\text{A}_2}\Big)+\text{P}(\text{A}_3)\cdot\text{P}\Big(\frac{\text{E}}{\text{A}_3}\Big)}$
$=\frac{\frac{4}{10}\cdot\frac{40}{100}}{\frac{4}{10}\cdot\frac{55}{100}+\frac{4}{10}\cdot\frac{40}{100}+\frac{2}{10}\cdot\frac{65}{100}}$
$=\frac{16}{51}$
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Question 805 Marks
The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs.
  1. None
  2. Not more than one
  3. More than one
  4. At least one
will fuse after 150 days of use.
Answer
Let X represent the number of bulbs that will fuse after 150 days of use in an experiment of 5 trials. The trials are Bernoulli trials. It is given that, p = 0.05 $\therefore\ \text{q}=1-\text{p}=1-0.05=0.95$ X has a binomial distribution with n = 5 and p = 0.05 $\therefore\ \text{p}(\text{X=x})=\ ^\text{n}\text{C}_\text{x}\text{q}^\text{n-x}\text{p}^\text{x},\ \text{where x}=1,\ 2,\ ...\text{n}$ $=\ ^5\text{C}_\text{x}(0.95)^{5-\text{x}}.(0.05)^\text{x}$
  1. P(none) = P(X = 0)
$=\ ^5\text{C}_{0}(0.95)^{5}.(0.05)^{0}$
$=1\times(0.95)^{5}$
$=(0.95)^{5}$
  1. P(not more than one) = P(X ≤ 1)
$\text{P}(\text{X}=0)+\text{P}(\text{X}=1)$
$=\ ^5\text{C}_0(0.95)^{5}\times(0.05)^{0}+\ ^5\text{C}_1(0.95)^{4}\times(0.05)^{1}$
$=1\times(0.95)^{5}+5\times(0.95)^{4}\times (0.05)$
$=(0.95)^{5}+(0.25)(0.95)^4$
$=(0.95)^{4}[0.95+0.25]$
$=(0.95)^{4}\times1.2$
  1. P(more than 1) = P(X > 1)
$=1-\text{P}(\text{X}\leq1)$
= 1 - P(not more than 1)
$=1-(0.95)^4\times1.2$
  1. P(at least one) = P(X ≥ 1)
$=1-\text{P}(\text{X}<1)$
$=1-\text{P}(\text{X}=0)$
$=1-\ ^5\text{C}_0(0.95)^5\times(0.05)^0$
$=1-1\times(0.95)^5$
$=1-(0.95)^5$
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Question 815 Marks
The contents of three bags $I, II$ and $III$ are as follows:
Bag $I : 1$ white, $2$ black and $3$ red balls,
Bag $II : 2$ white, $1$ black and $1$ red ball;
Bag $III : 4$ white, $5$ black and $3$ red balls.
A bag is chosen at random and two balls are drawn. What is the probability that the balls are white and red?
Answer
A white ball and a red ball can be drawn in three mutually exclusice ways:
  1. Selecting bag $I$ and then drawing a white and a red ball from it.
  2. Selecting bag $II$ and then drawing a white and a red ball from it.
  3. Selecting bag $III$ and then drawing a white and a red ball from it.
Let $E_1, E_2$ and $A$ be the events as defined below;
$E_1 =$ Selecting bag $I$
$E_2 =$ Selecting bag $II$
$E_3 =$ Selecting bag $III$
$A =$ Drawing $A$ white and a red ball
It is given that one of the bags is selected randomly.
$\therefore\ \text{P}(\text{E}_1)=\frac{1}{3}$
$\text{P}(\text{E}_2)=\frac{1}{3}$
$\text{P}(\text{E}_3)=\frac{1}{3}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{^{1}\text{C}_1\times ^{3}\text{C}_1}{^{6}\text{C}_2}=\frac{3}{15}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{^{2}\text{C}_1\times ^{1}\text{C}_1}{^{4}\text{C}_2}=\frac{2}{6}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=\frac{^{4}\text{C}_1\times ^{3}\text{C}_1}{^{12}\text{C}_2}=\frac{12}{66}$
Using the law of total probability, we get
Required probability $=\text{P(A)}=\text{P}(\text{E})_1\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)$
$=\frac{1}{3}\times\frac{3}{15}+\frac{1}{3}\times\frac{2}{6}+\frac{1}{3}\times\frac{12}{66}$
$=\frac{1}{15}+\frac{1}{9}+\frac{2}{33}$
$=\frac{33+55+30}{495}$
$=\frac{118}{495}$
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Question 825 Marks
An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?
Answer
Let E and F denote respectively the events that first and second ball drawn are black. We have to find $\text{P}(\text{E}\cap\text{F})$ or P(EF)
Now P(E) = P (black ball in first draw) $=\frac{10}{15}$
Also given that the first ball drawn is black, i.e, event E has occurred, now there are 9 black balls and five white balls left in the urn. Therefore, the probability that the second ball drawn is black, given that the ball in the first draw is black, is nothing but the conditional probability of F given that E has occurred.
i.e., $\text{P}(\text{F}|\text{E})=\frac{9}{14}$
By multiplication rule of probability, we have
$\text{P}(\text{E}\cap\text{F})=\text{P}(\text{E})\text{ P}(\text{F}|\text{E})$
$=\frac{10}{15}\times\frac{9}{14}=\frac{3}{7}$
Multiplication rule of probability for more than two events if E,F and G are three events of sample space, we have
$\text{P}(\text{E}\cap\text{F}\cap\text{G})=\text{P}(\text{E})\text{ P}(\text{F}|\text{E}) \text{ P}(\text{G}) (\text{G}\cap\text{F})= \text{P}(\text{E})\text{ P}(\text{F}|\text{E}) \text{ P}(\text{G}|\text{EF})$
Similarly, the multiplication rule of probability can be extended for four or more events.
The following example illustrates the extension of multiplication rule of probability for three events.
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Question 835 Marks
There are three coins. One is two-headed coin (having head on both faces), another is biased coin that comes up heads 75% of the times and third is also a biased coin that comes up tail 40% of the times. One of the three coins is chosen at random and tossed, and it shows heads. What is the probability that it was the two-headed coin?
Answer
A be the event of choosing two - headed coin,
B be the event of choosing a biased coin that comes up head 75% of the times,
C be the event of choosing a biased coin that comes up tail 40% of the times and
E be the event of getting a head.
Now,
$\text{P(A)}=\text{P(B)}=\text{P(C)}=\frac{1}{3}$ and
$\text{P}(\text{E}|\text{A})=1,\text{P}(\text{E}|\text{B})=75\%=\frac{75}{100}=\frac{3}{4}$ and $\text{P}(\text{E}|\text{C})=60\%=\frac{60}{100}=\frac{3}{5}$
So, using Bayes' theorem, we get
P (the head shown was of two - headed coin) = P(A|E)
$=\frac{\text{P(A)}\times\text{P}(\text{E}|\text{A})}{\text{P(A)}\times\text{P}(\text{E}|\text{A})+\text{P(B)}\times(\text{E}|\text{B})+\text{P(C)}\times\text{P}(\text{E}|\text{C})}$
$=\frac{\Big(\frac{1}{3}\times1\Big)}{\Big(\frac{1}{3}\times1+\frac{1}{3}\times\frac{3}{4}+\frac{1}{3}\times\frac{3}{5}\Big)}$
$=\frac{\Big(\frac{1}{3}\Big)}{\Big(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\Big)}$
$=\frac{\Big(\frac{1}{3}\Big)}{\Big(\frac{20+15+12}{60}\Big)}$
$=\frac{\Big(\frac{4}{3}\Big)}{\Big(\frac{47}{60}\Big)}$
$=\frac{60}{3\times47}$
$=\frac{20}{47}$
So, the probability that the head shown was of a two-headed coin is $=\frac{20}{47}$.
Disclaimer: The answer given in the book is incorrect. The same has been corrected here.
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Question 845 Marks
The probabilities of two students A and B coming to the school in time are $\frac{3}{7}$ and $\frac{5}{7}$ respectively. Assuming that the events, 'A coming in time' and 'B coming in time' are independent, find the probability of only one of them coming to the school in time. Write at least one advantage of coming to school in time.
Answer
Given that the events 'A coming in time' and 'B coming in time' are independent.
Let 'A' denote the event of 'A coming in time'.
Then, $'\overline{\text{A}'}$ denotes the complementary event of A.
Similarly define B and $\overline{\text{B}}$.
P(Only one coming in time) $=\text{P}(\text{A}\cap\overline{\text{B}})+\text{P}(\overline{\text{A}}\cap\text{B})$
$=\text{P(A)}\times\text{P}(\overline{\text{B}})+\text{P}(\overline{\text{A}})\times\text{P(B)}\ ......$
(Since A and B are independent events)
$=\frac{3}{7}\times\frac{2}{7}+\frac{4}{7}\times\frac{5}{7}=\frac{6}{49}+\frac{20}{49}=\frac{26}{49}$
The advantage of coming to school in time is that you will not miss any part of the lecture and will be able to learn more.
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Question 855 Marks
The probability of a shooter hitting a target is $\frac{3}{4}.$ How many minimum number of times must he/she fire so that the probability of hitting the target at least once is more than 0.99?
Answer
Let the shooter fire n times.
n fires are bernoulli trials.
In each trial, p = probability of hitting the target $=\frac{3}{4}$
And q = probability of not hitting the target $=1-\frac{3}{4}=\frac{1}{4}$
Then, $\text{P(X = x})=\text{ }^{\text{n}}\text{c}_{\text{x}}\text{q}^{\text{n}-\text{x}}\text{p}^{\text{x}}=\text{ }^{\text{n}}\text{c}_{\text{x}}\big(\frac{1}{4}\big)^{\text{n}-\text{x}}\big(\frac{3}{4}\big)^{\text{x}}=\text{ }^{\text{n}}\text{c}_{\text{x}}\frac{3^{\text{x}}}{4^{\text{n}}}$
Now, given that
P(hitting the target atleast once) > 0.99
i.e. $\text{P(X}\geq1)>0.99$
$\Rightarrow1-\text{P(X}=0)>0.99$
$\Rightarrow1-\text{ }^{\text{n}}\text{c}_0\frac{1}{4^{\text{n}}}>0.99$
$\Rightarrow\text{ }^{\text{n}}\text{c}_0\frac{1}{4^{\text{n}}}<0.01$
$\Rightarrow\frac{1}{4^{\text{n}}}<0.01$
$\Rightarrow4^{\text{n}}>\frac{1}{0.01}=100$
The minimum value of n to satisfy this inequality is 4
Thus, the shooter must fire 4 times.
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Question 865 Marks
A fair die is tossed. Let X denote 1 or 3 according as an odd or an even number appears. Find the probability distribution, mean and variance of X.
Answer
Let X be 1 for the appearance of odd numbers 1, 3 or 5 on the die. Then, $\text{P}(\text{X}=1)=\frac{3}{6}=\frac{1}{2}$ Let X be 3 for the appearance of even numbers 2, 4 or 6 on the die. Then, $\text{P}(\text{X}=3)=\frac{3}{6}=\frac{1}{2}$ Thus, the probability distribution of X is given by
$\text{X}$ $\text{P}(\text{X})$
$1$ $\frac{1}{2}$
$3$ $\frac{1}{2}$
Computation of mean and variance
$\text{x}_\text{i}$ $\text{p}_\text{i}$ $\text{p}_\text{i}\text{x}_\text{i}$ $\text{p}_\text{i}\text{x}_\text{i}^2$
$2$ $\frac{1}{2}$ $\frac{1}{2}$ $\frac{1}{2}$
$4$ $\frac{1}{2}$ $\frac{3}{2}$ $\frac{9}{2}$
    $\sum\text{x}_\text{i}\text{p}_\text{i}=2$ $\sum\text{x}_\text{i}\text{p}_\text{i}^2=5$
Mean $=\sum\text{x}_\text{i}\text{p}_\text{i}=2$ Variance $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-(\text{Mean})^2$ $=5-4$ $=1$
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Question 875 Marks
A card is drawn from a pack of 52 cards so that each card is equally likely to be selected. In which of the following cases are the events A and B independent?
A = The card drawn is a king or queen,
B = the card drawn is a queen or jack.
Answer
A card is drawn from 52 cards
It has 4 kings, 4 queen, 4 jack
A = The card drawn is a king ir a queen
$\text{P(A)}=\frac{4+4}{52}$
$=\frac{8}{52}$
$\text{P(A)}=\frac{2}{13}$
B = the card drawn is a queen or a jack
$\text{P(B)}=\frac{4+4}{52}$
$=\frac{8}{52}$
$=\frac{2}{13}$
$\text{A}\cap\text{B}=$ The card drawn is a queen
$\text{P}(\text{A}\cap\text{B})=\frac{4}{52}$
$=\frac{1}{13}$
$\text{P(A)}\text{ P(B)}=\frac{2}{13}\times\frac{2}{13}$
$=\frac{4}{169}$
$\text{P(A)}\text{ P(B)}\neq\text{P}(\text{A}\cap\text{B})$
Hence, A and B are not independent.
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Question 885 Marks
The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting at least twice?
Answer
Let hitting the target be a success in a shoot.
We have,
p = probability of hitting the target $=0.25=\frac{1}{4}$
Also, $\text{q}=1-\text{p}=1=\frac{1}{4}=\frac{3}{4}$
Let X denote the number of success in a sample of 7 trils. then,
X follows binomial distribution with parameters n = 7 and $\text{p}=\frac{1}{4}$
$\therefore\text{P(X = r})=\text{ }^7\text{C}_{\text{r}}\text{p}^{\text{r}}\text{q}^{(7-\text{r})}=\text{ }^7\text{C}_{\text{r}}\big(\frac{1}{4}\big)^{\text{r}}\big(\frac{3}{4}\big)^{(7-\text{r})}=\frac{\text{ }^7\text{C}_{\text{r}}3^{(7-\text{r})}}{4^7},$ where r = 0, 1, 2, 3, 4, 5
Now,
Required probability $=\text{P(X}\geq2)$
$=1-\big[\text{P(X}=0)+\text{P(X}=1)\big]$
$=1-\Big[\frac{\text{ }^7\text{C}_03^7}{4^7}+\frac{\text{ }^7\text{C}_13^6}{4^7}\Big]$
$=1-\Big[\frac{2187}{16384}+\frac{5103}{16384}\Big]$
$=1-\frac{7290}{16384}$
$=\frac{9094}{16384}$
$=\frac{4547}{8192}$
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Question 895 Marks
Bag $I$ contains $3$ red and $4$ black balls and Bag $II$ contains $4$ red and $5$ black balls. One ball is transferred from Bag $I$ to Bag $II$ and then a ball is drawn from Bag $II.$ The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.
Answer
Bag $I$ contains $3$ red and $4$ black balls.
Bag $II$ contain $4$ red and $5$ black balls.
Let $E_1 :$ Evenr that a red ball is drawn from bag $I$
$E_2 :$ Event that a black ball is drawn from bag $I$
$\therefore\ \text{P}(\text{E}_1)=\frac{3}{7},\ \text{P}(\text{E}_2)=\frac{4}{7}$
After transferring a red ball from bag $I$ to bag $II,$ the bag $II$ will have $5$ red and $5$ black balls.
Let $A$ be the event of drawing red ball
$\therefore\ \text{P}(\text{A|E}_1)=\frac{5}{10}=\frac{1}{2}$
Further when a black ball is transferred from bag $I$ to bag $II,$ it will contain $4$ red and $6$ black balls.
$\text{P}(\text{A|E}_2)=\frac{4}{10}=\frac{2}{5}$
$\text{Required probability}\ =\text{P}(\text{E}_2|\text{A})=\frac{\text{P}(\text{E}_2)\text{P}(\text{A|E}_2)}{\text{P}(\text{E}_1)\text{P}(\text{A|E}_1)+\text{P}(\text{E}_2)\text{P}(\text{A|E}_2)}$
$=\frac{\frac{4}{7}\times\frac{2}{5}}{\frac{3}{7}\times\frac{1}{2}+\frac{4}{7}\times\frac{2}{5}}$
$=\frac{\frac{8}{35}}{\frac{13}{14}+\frac{8}{35}}$
$=\frac{\frac{8}{35}}{\frac{15+16}{70}}$
$=\frac{8}{35}\times\frac{70}{31}$
$=\frac{16}{31.}$
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Question 905 Marks
A card is drawn from a pack of 52 cards so that each card is equally likely to be selected. In which of the following cases are the events A and B independent?
A = the card drawn is black,
B = the card drawn is a king.
Answer
A card is drawn from pack of 52 cards
There are 26 black and four kings in which 2 kings are black.
A = the card drawn is black
$\text{P(A)}=\frac{26}{52}$
$\text{P(A)}=\frac{1}{2}$
B = the card drawn is a king
$\text{P(B)}=\frac{4}{52}$
$=\frac{1}{13}$
$\text{A}\cap\text{B}=$ The card drawn is a black king
$\text{P}(\text{A}\cap\text{B})=\frac{2}{52}=\frac{1}{26}$
$\text{P(A) }\text{P(B)}=\frac{1}{2}\times\frac{1}{13}$
$=\frac{1}{26}$
$\text{P(A) }\text{P(B)}=\text{P}(\text{A}\cap\text{B})$
So, A and B are independent events.
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Question 915 Marks
A coin is tossed thrice and all the eight outcomes are assumed equally likely. In which of the following cases are the following events A and B are independent?
A = The number of heads is odd,
B = The number of tails is odd.
Answer
Sample space for a coin thrown thrice is
= {HHT, HTT, THT, TTT, HHH, HTH, THH, TTH}
A = the number of head is odd
A = {HTT, THT, TTH, HHH}
B = the number if tails is odd
B = {THH, HTH, HHT, TTT}
$\text{A}\cap\text{B}=\{\}=\phi$
$\text{P(A)}=\frac{4}{8}=\frac{1}{2}$
$\text{P(B)}=\frac{4}{8}=\frac{1}{2}$
$\text{P}(\text{A}\cap\text{B})=\frac{0}{8}=0$
$\text{P(A)}.\text{P(B)}=\frac{1}{2}\times\frac{1}{2}$
$=\frac{1}{4}$
$\text{P(A)}.\text{P(B)}\neq\text{P}(\text{A}\cap\text{B})$
So, A and B are not independent events.
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Question 925 Marks
From a lot of 30 bulbs that includes 6 defective bulbs, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.
Answer
Let X denote the number of defective bulls in a sample of 4 bulbs drawn successively with replacement.
Then, X follows a binomial distribution with the following parameters: n = 4,
$\text{p}=\frac{6}{30}=\frac{1}{5}$ and $\text{q}=\frac{4}{5}$
Then, the distribution is given by
$\text{P(X = r})=\text{ }^4\text{C}_{\text{r}}\big(\frac{1}{5}\big)^{\text{r}}\big(\frac{4}{5}\big)^{4-\text{r}}.\text{r}=0,1,2,3,4$
$\text{P(X = } 0)=\big(\frac{4}{5}\big)^4$
$=\frac{256}{625}$
$\text{P(X}=1)=4\big(\frac{1}{5}\big)^1\big(\frac{4}{5}\big)^3$
$=\frac{256}{625}$
$\text{P(X}=2)=6\big(\frac{1}{5}\big)^2\big(\frac{4}{5}\big)^2$
$=\frac{96}{625}$
$\text{P(X}=3)=4\big(\frac{1}{5}\big)^3\big(\frac{4}{5}\big)^1$
$=\frac{16}{625}$
$\text{P(X}=4)=\big(\frac{1}{5}\big)^4$
$=\frac{1}{625}$
$\text{X}$ $1$ $2$ $3$ $4$ $5$
$1\text{P(X)}$ $\frac{256}{625}$ $\frac{256}{625}$ $\frac{96}{625}$ $\frac{16}{625}$ $\frac{1}{625}$
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Question 935 Marks
A fair coin is tossed four times. Let X denote the number of heads occuring. Find the probability distribution, mean and variance of X.
Answer
We know that, in a toss of coin. $\text{P(T)}=\frac{1}{2},\text{P(H)}=\frac{1}{2}$ Let X denote the number of accuring head in four throws of a coins. So, X can take values from X = 0, 1, 2, 3, 4 $\text{P(X=0)}=\text{P(T)}\text{P(T)}\text{P(T)}\text{P(T)}$ $=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$ $=\frac{1}{16}$ $\text{P(X=1)}=\text{P(H)}\text{P(T)}\text{P(T)}\text{P(T)}\times{^{4}}\text{C}_1$ $=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times4$ $=\frac{4}{16}$ $\text{P(X=2)}=\text{P(H)}\text{P(H)}\text{P(T)}\text{P(T)}\times{^{4}}\text{C}_2$ $=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times6$ $=\frac{6}{16}$ $\text{P(X=3)}=\text{P(H)}\text{P(H)}\text{P(H)}\text{P(T)}\times{^{4}}\text{C}_3$ $=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times4$ $=\frac{4}{16}$ $\text{P(X=4)}=\text{P(H)}\text{P(H)}\text{P(H)}\text{P(H)}$ $=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$ $=\frac{1}{16}$ So,
$\text{x}_\text{i}$ $\text{p}_\text{i}$ $\text{x}_\text{i}\text{p}_\text{i}$ $\text{x}_\text{i}^2\text{p}_\text{i}$
$0$ $\frac{1}{16}$ $0$ $0$
$1$ $\frac{4}{16}$ $\frac{4}{16}$ $\frac{4}{16}$
$2$ $\frac{6}{16}$ $\frac{12}{16}$ $\frac{24}{16}$
$3$ $\frac{4}{16}$ $\frac{12}{16}$ $\frac{36}{16}$
$4$ $\frac{1}{16}$ $\frac{4}{16}$ $\frac{16}{16}$
    $\sum\text{xp}=2$ $\sum\text{x}^2\text{p}=5$
Mean $=\sum\text{xp}=2$ Variance $=\sum\text{x}^2\text{p}-(\text{mean})^2$ $=5-(2)^2=1$ Probability distribution is
$\text{x}:$ $0$ $1$ $2$ $3$ $4$
$\text{p(x)}:$ $\frac{1}{16}$ $\frac{4}{16}$ $\frac{6}{16}$ $\frac{4}{16}$ $\frac{1}{16}$
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Question 945 Marks
An insurance company insured $2000$ scooters and $3000$ motorcycles. The probability of an accident involving a scooter is $0.01$ and that of a motorcy is $0.02.$ An insured vehicle met with an accident. Find the probability that the accidented vehicle was a motorcycle.
Answer
Let $E_1, E_2$ and $A$ be events ar:
$E_1 =$ Vehilcle is scooter
$E_2 =$ Vehicle is motorcycle
$A =$ An insured met with scooter
$\text{P}(\text{E}_1)=\frac{2000}{5000}=\frac{2}{5}$
$\text{P}(\text{E}_2)=\frac{3000}{5000}=\frac{3}{5}$
$P(A|E_1) = P($Accident of scooter$)$
$= 0.01$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)= P($Accident of motorcycle$)$
$= 0.02$
To find, $P($Accident vehicle was motorcycle$) =\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big)$
By Baye's theorem,
$\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big)=\frac{\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{3}{5}\times\frac{2}{100}}{\frac{2}{5}\times\frac{1}{100}+\frac{3}{5}\times\frac{2}{100}}$
$=\frac{\frac{6}{500}}{\frac{2}{500}+\frac{6}{500}}$
$=\frac{6}{8}$
$=\frac{3}{4}$
Required probability $=\frac{3}{4}$
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Question 955 Marks
Bag $I$ contains $3$ black and $2$ white balls, Bag $II$ contains $2$ black and $4$ white balls. A bag and a ball is selected at random. Determine the probability of selecting a black ball.
Answer
Bag $I = \{3B, 2W\},$ Bag $II = \{2B, W\}$
Let $E_1=$ Event that bag $I$ is selected
$E_2=$ Event that bag $II$ is selected
And $E =$ Event that a black ball is selected
$\Rightarrow\text{P}(\text{E}_1)=\frac{1}{2},\text{P}(\text{E}_2)=\frac{1}{2},$ 
$\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)=\frac{3}{5},\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)=\frac{2}{6}=\frac{1}{3}$
$\therefore\text{P}(\text{E})=\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)$
$=\frac{1}{2}\cdot\frac{3}{5}+\frac{1}{2}\cdot\frac{2}{6}$
$=\frac{3}{10}+\frac{2}{12}$
$=\frac{18+10}{60}$
$=\frac{28}{60}$
$=\frac{7}{15}$
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Question 965 Marks
A bag contains 25 tickets, numbered from 1 to 25. A ticket is drawn and then another ticket is drawn without replacement. Find the probability that both tickets will show even numbers.
Answer
Tickets are numbered from 1 to 25
⇒ Total number of tickets = 25
Number of tickets with even numbers on it
= 12 {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24}
A = first ticket with even number
B = second ticket with even number
P (Both tickets will show even number, without replacement)
$=\text{P}(\text{A})\ \text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{12}{25}\times\frac{11}{24}$
$=\frac{11}{50}$
Required probability $=\frac{11}{50}$
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Question 975 Marks
A bag contains 3 red and 5 black balls and a second bag contains 6 red and 4 black balls. A ball is drawn from each bag. Find the probability that one is red and the other is black.
Answer
There are two bags.
Bag (1) contain 3 red and 5 black balls
Bag (2) contain 6 red and 4 black balls
P (One red ball from bag 1) $=\frac{3}{8}$
$\text{P}(\text{R}_1)=\frac{3}{8}$
P (One black ball from bag 1) $=\frac{5}{8}$
$\text{P}(\text{B}_1)=\frac{5}{8}$
P (One red ball from bag 1) $=\frac{6}{10}$
$\text{P}(\text{R}_2)=\frac{3}{5}$
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Question 985 Marks
A bag contains 10 balls, each marked with one of the digits from 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?
Answer
Let p denote the probability of getting a ball market with 0. So
$\text{p}=\frac{1}{10}$ [Since balls are market with 0, 1, 2, 3, 4, 5, 6, 7, 8, 8, 9]
$\text{q}=1-\frac{1}{10}$ [Since p + q = 1]
$\text{q}=\frac{9}{10}$
Let X denote the variable presenting the number of balls marked with 0 out of four balls drawn. probability of drawing r balls out of n balls that are marked 0 is given by
$\text{P}(\text{X = r})=\text{ }^\text{n}\text{c}_\text{r}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$=\text{ }^{4}\text{c}_\text{r}\big(\frac{1}{10}\big)^\text{r}\big(\frac{9}{10}\big)^{4-\text{r}}\dots(1)$
Probability of getting none balls marked with 0
$=\text{P}(\text{X}=0)$
$=\text{ }^4\text{c}_0\big(\frac{1}{10}\big)^0\big(\frac{9}{10}\big)^{4-0}$
$=1.1.\big(\frac{9}{10}\big)^4$
$=\big(\frac{9}{10}\big)^4$
Probability of getting nine balls marked with $0=\big(\frac{9}{10}\big)^4$
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Question 995 Marks
In a large bulk of items, $5$ percent of the items are defective. What is the probability that a sample of $10$ items will include not more than one defective item?
Answer
Let $X$ denote the number of defective items in a sample of $10$ items.
$X$ follows a binomial distribution with $n = 10;$
$P =$ Probability of detective items $= 5\% = 0.05; q = 1- p = 0.95$
$P(X = r)=Cr10(0.05)r(0.95) 10 - r$
$P(x=r) = ^{10}C_r(0.05)^{r }(0.95)^{10-r}$
Probability $($sample of $10$ items will include not more than one defective item$)=P(X \leq 1)$
$= P(X = 0) + P(X = 1)$
$= ^{10}C_0(0.05)^{0 }(0.95)^{10-0 }+ ^{10}C_1(0.05)^{1 }(0.95)^{10-1}$
$= (0.95)^{9 }(0.95+0.5)$
$= 1.45(0.95)^9$
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Question 1005 Marks
A letter is known to have come either from $\text{LONDON}$ or $\text{CLIFTON}$. On the envelope just two consecutive letters $\text{ON}$ are visible. What is the probability that the letter has come from$, \text{LONDON}$.
Answer
Consider events $E_1, E_2$ and $A$ events As:
$E_1 =$ Letters come from $\text{LONDON}$
$E_2 =$ Letters come from $\text{CLIFTON}$
$E_3 =$ Two consecutive letters visible on the envelope are on
$\text{P}(\text{E}_1)=\frac{1}{2}$
$\text{P}(\text{E}_2)=\frac{1}{2}$
$[$Since letters came either from $\text{LONDON}$ or $\text{CLIFTON}]$
$P(A | E_1) = P($Two consecutive letters $\text{ON}$ from $\text{LONDON})$
$=\frac{2}{5}$
$[$Since $\text{LONDON}$ has $2 - \text{ON}$ and $5$ letters we consider one $'ON\ '$ as one letter$]$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}\ ($Two consecutive letters On from $\text{CLIFTON})$
$=\frac{1}{6}$
$[$Since $\text{CLIFTON}$ has one $'\text{ON}\ '$ nad $6$ letters considering $\text{ON}$ as one letter$]$
To find$, P (\text{ON}$ visible are from $\text{LONDON}) \text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big).$
By baye's theorem,
$\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{1}{2}\times\frac{2}{5}}{\frac{1}{2}\times\frac{2}{5}+\frac{1}{2}\times\frac{1}{6}}$
$=\frac{\frac{2}{10}}{\frac{2}{10}+\frac{1}{12}}$
$=\frac{2}{10}\times\frac{60}{17}$
$=\frac{12}{17}$
$\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{12}{17}$
Required probability $=\frac{12}{17}$
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5 Marks Questions - Page 2 - MATHS STD 12 Science Questions - Vidyadip