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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
A card from a pack of $52$ cards is lost. From the remaining cards of the pack, two cards are drawn at random and are found to both clubs. Find the probability of the lost card being of clubs.
Answer
Let $E_1$ be the event that lost card is that of clubs.
$E_2$ be event that lost card is not of clubs.
A: Two cards of clubs are drawn from remaining cards.
$\text{P(E}_{1})=\frac{1}{4},\text{P(E}_{2})=\frac{3}{4}$
$\text{P(A/E}_{1})=\text{12C}_{2}/{\text{51C}_{2}}=\frac{12\times11}{51\times50}$
$\text{P(A/E}_{1})=\text{12C}_{2}/{\text{51C}_{2}}=\frac{2\times11}{17\times25}=\frac{22}{425}$
$\text{P(A/E}_{2})=\text{13C}_{2}/{\text{51C}_{2}}=\frac{13\times12}{51\times50}$
$\text{P(A/E}_{2})=\text{13C}_{2}/{\text{51C}_{2}}=\frac{13\times2}{17\times25}=\frac{26}{425}$
$\text{P(E}_{1}/\text{A})=\frac{\text{P(A/E}_{1})\cdot\text{P(E}_{1})}{\sum\text{P(E}_{1}\cdot\text{P(A/E}_{1})}$
$=\frac{\frac{22}{425}\times\frac{1}{4}}{\frac{22}{425}\times\frac{1}{4}+\frac{26}{425}\times\frac{3}{4}}=\frac{11}{50}$.
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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
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 605 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 615 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 625 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 635 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 645 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 655 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 665 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 675 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 685 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 695 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 705 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 715 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 725 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 735 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 745 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 755 Marks
In a multiple-choice examination with three possible answers for each of the five questions out of which only one is correct, what is the probability that a candidate would get four or more correct answers just by guessing?
Answer
The repeated guessing of correct answers form multiple choice questions are bernoulli trials. Let X represent the number of correct answers by guessing in the set of 5 multiple choice questions.Probability of getting a correct answer is, $\text{p}=\frac{1}{3}$
$\therefore\text{q}=1-\text{p}=1-\frac{1}{3}=\frac{2}{3}$
Clearly, X has a binomial distribution with $\text{n = 5}$ and $\text{p}=\frac{1}{3}$
$\therefore\text{P(X = x)}=\text{ }^{\text{n}}\text{C}_{\text{x}}\text{q}^{\text{n}-\text{x}}\text{p}^{\text{x}}$
$=\text{ }^5\text{C}_{\text{x}}\big(\frac{2}{3}\big)^{5-\text{x}}.\big(\frac{1}{3}\big)^{\text{x}}$
P(guessing more than 4 correct answer) $=\text{P(X}\geq4)$
$=\text{P(X}=4)+\text{P(X}=5)$
$=\text{ }^5\text{C}_4\big(\frac{2}{3}\big).\big(\frac{1}{3}\big)^4+\text{ }^5\text{C}_5\big(\frac{1}{3}\big)^5$
$=5.\frac{2}{3}.\frac{1}{81}+1.\frac{1}{243}$
$=\frac{10}{243}+\frac{1}{243}$
$=\frac{11}{243}$
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Question 765 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 775 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 785 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 795 Marks
Given that the two numbers appearing on throwing two dice are different. Find the probability of the event 'the sum of numbers on the dice is 4'.
Answer
A = Two numbers on two dice are different
= {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
B = Sum of numbers on the dice is 4
B = {(1, 3), (2, 2), (3, 1)}
$\text{A}\cap\text{B}=\{(1,3),(3,1)\}$
Required probability $=\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{A})}$
$=\frac{2}{30}$
Required probability $=\frac{1}{15}$
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Question 805 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 815 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 825 Marks
A letter is known to have come either from LONDON or CLIFTON. On the envelope just two consecutive letters ON are visible. What is the probability that the letter has come from,
LONDON.
Answer
Consider events $E_1, E_2$ and A events As:
$E_1$ = Letters come from LONDON
$E_2$ = Letters come from 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 LONDON or CLIFTON]
$P(A | E_1)$ = P(Two consecutive letters ON from LONDON)
$=\frac{2}{5}$
[Since LONDON has 2 - 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 CLIFTON)
$=\frac{1}{6}$
[Since CLIFTON has one 'ON' nad 6 letters considering ON as one letter]
To find, P (ON visible are from 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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Question 835 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 845 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 855 Marks
If $\text{P}(\text{not B})=0.65, \text{P}(\text{A}\cup\text{B})=0.85$, and A and B are independent events, then find P(A).
Answer
$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
Since A, B are independent
$\therefore\text{P}(\text{A}\cap\text{B})=\text{P(A) }\text{P(A)}$
Also, $\text{P(noy B)} = 0.65 \Rightarrow \text{P(B)} = 0.35$
Hence, we have
$0.85 = \text{P(A)} + 0.35 - \text{P(A)} (0.35)$
$\Rightarrow 0.8 = \text{P(A)} [1 - 0.35]$
$\Rightarrow \frac{0.5}{.65}=\text{P(A)}$
$\Rightarrow \text{P(A)} = 0.77$
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Question 865 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)$=P(X ≤ 1)$
$= P(X = 0) + P(X = 1)$
$= ^{10}C_0(0.05)^0(0.95)^{10-0}+ ^{10}C_1(0.05)^1(0.95)^{10-1}$
$= (0.95)^9(0.95+0.5)$
$= 1.45(0.95)^9​​​​​​​$
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Question 875 Marks
The bag A contains 8 white and 7 black balls while the bag B contains $5$ white and $4$ black balls. One ball is randomly picked up from the bag A and mixed up with the balls in bag B. Then a ball is randomly drawn out from it. Find the probability that ball drawn is white.
Answer
Bag A contains 8 white and 7 black balls
Bag B contains 5 white and 4 black balls
Transfer can be done in two ways:
I - A white ball is transferred from bag A to bag B and then onw white ball is drawn from bag B.
II - A black ball is transferred fron bag A to bag, then one white ball is drawn from bag B.
Let$ E_1, E_2$ and A be events as:
$E_1$ = One white ball from bag A
$E_2$ = one black ball from bag A
A = One white ball from bag B
$\text{P}(\text{E}_1)=\frac{8}{15}$
$\text{P}(\text{E}_2)=\frac{7}{15}$
$\text{P}(\text{A}|\text{E}_1)=\frac{6}{10}$
[Since $E_1$ has increased white balls in bag B]
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{5}{10}$
[Since $E_2$ has increased black ball in bag B]
By law of total probability,
$\text{P(A)}=\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)$
$=\frac{8}{15}\times\frac{6}{10}+\frac{7}{15}\times\frac{5}{10}$
$=\frac{48}{150}+\frac{35}{150}$
$=\frac{83}{150}$
Required probability $=\frac{83}{150}$
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Question 885 Marks
By examining the chest X ray, the probability that TB is detected when a person is actually suffering is $0.99$. The probability of an healthy person diagnosed to have TB is $0.001$. In a certain city, $1$ in $1000$ people suffers from TB. A person is selected at random and is diagnosed to have TB. What is the probability that he actually has TB?
Answer
Let $E_1$ = Event that person has TB
$E_2$ = Event that the person does not have TB
$E$ = Event that the person is diagnosed to have TB
$\therefore\text{P}(\text{E}_1)=\frac{1}{1000}=0.001,$
$\text{P}(\text{E}_2)=\frac{999}{1000}=0.999$
And $\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)=0.099$ and $\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)=0.001$
$\therefore\text{P}\Big(\frac{\text{E}_1}{\text{E}}\Big)=\frac{\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)}$
$=\frac{0.001\times0.99}{0.001\times0.99+0.999\times0.001}$
$=\frac{0.000990}{0.000990+0.000999}$
$=\frac{990}{1989}=\frac{110}{221}$
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Question 895 Marks
Assume that the chances of a patient having a heart attack is $40\%$. It is also assumed that a meditation and yoga course reduce the risk of heart attack by $30\%$ and prescription of certain drug reduces its chances by $25\%$. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?
Answer
Let $E_1$ and $E_2$ be the events
$E_1$ : Treatment of yoga and meditation
$E_2$ : Treatment of prescription of certain drugs
$\therefore\ \text{P}(\text{E}_1)=\frac{1}{2},\ \text{P}(\text{E}_2)=\frac{1}{2}$
Let A denotes that a person has heart risk attack
$\therefore\ \text{P}(\text{A})=40\%=0.40$
Yoga and mediation reduces the heart risk by 30%
i. e. inspite of getting yoga and meditation treatment heart risk is 70% of the 0.40
$\therefore\ \text{P}(\text{A|}\text{E}_1)=0.40\times0.70=0.28$
Drug prescription reduces the heart risk by 25%
Even after adopting the drug prescription heart risk is 75% of the 0.40
$\therefore\ \text{P}(\text{A|E}_2)=0.40\times0.75=0.30$
$\therefore\ \text{P}(\text{E}_1|\text{A})=\frac{\text{P}(\text{E}_1)\text{P}(\text{A|E}_1)}{\text{P}(\text{E}_1)\text{P}(\text{A|E}_1)+\text{P}(\text{E}_2)\text{P}(\text{A|E}_2)}$
$=\frac{\frac{1}{2}\times0.28}{\frac{1}{2}\times0.28+0.30\frac{1}{2}}=\frac{0.28}{0.28+0.30}=\frac{28}{58}=\frac{14}{29}$
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Question 905 Marks
A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of the number of successes.
Answer
Let p be the probability of getting a doublet in a throw of a pair of dice, so
$\text{p}=\frac{6}{36}$ [Since (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)]
$=\frac{1}{6}$
$\text{q}=1-\frac{1}{6}$ [Since p + q = 1]
$=\frac{5}{6}$
Let X denote the number of grtting doublets i.e. success out of 4 times. So, probility distribution is given by
$\text{X}$ $\text{P(X)}$
$0$ $\text{ }^4\text{C}_0\big(\frac{1}{6}\big)^0\big(\frac{5}{6}\big)^{4-0}=\big(\frac{5}{6}\big)^4$
$1$ $\text{ }^4\text{C}_1\big(\frac{1}{6}\big)^1\big(\frac{5}{6}\big)^{4-1}=4\big(\frac{1}{6}\big)\big(\frac{5}{6}\big)^3=\frac{2}{3}\big(\frac{5}{6}\big)^3$
$2$ $\text{ }^4\text{C}_2\big(\frac{1}{6}\big)^2\big(\frac{5}{6}\big)^{4-2}=\frac{4.3}{2}\big(\frac{1}{6}\big)^2\big(\frac{5}{6}\big)^2=\frac{25}{216}$
$3$ $\text{ }^4\text{C}_3\big(\frac{1}{6}\big)^3\big(\frac{5}{6}\big)^{4-3}=\frac{4.3}{2}\big(\frac{1}{6}\big)^3\big(\frac{5}{6}\big)=\frac{5}{324}$
$4$ $\text{ }^4\text{C}_4\big(\frac{1}{6}\big)^4\big(\frac{5}{6}\big)^{4-4}=\big(\frac{1}{6}\big)^4=\frac{1}{1296}$
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Question 915 Marks
Two cards are drawn simultaneously from a pack of 52 cards. Compute the mean and standard deviation of the number of kings.
Answer
Two cards are drawn simultaneously from a pack of 52 cards. Let X denotes the number of kings drawn. So, X = 0, 1, 2. P(X = 0) $=\frac{\text{}^{48}\text{C}_2}{\text{}^{52}\text{C}_2}$ $=\frac{1128}{1326}$ $=\frac{188}{221}$ P(X = 1) $=\frac{\text{}^{4}\text{C}_1\times\text{}^{48}\text{C}_1}{\text{}^{52}\text{C}_2}$ $=\frac{192}{1326}$ $=\frac{32}{221}$ P(X = 2) $=\frac{\text{}^{4}\text{C}_2}{\text{}^{52}\text{C}_2}$ $=\frac{6}{1326}$ $=\frac{1}{221}$ So,
$\text{x}_\text{i}$ $\text{p}_\text{i}$ $\text{x}_\text{i}\text{p}_\text{i}$ $\text{x}_\text{i}^2\text{p}_\text{i}$
$0$ $\frac{188}{221}$ $0$ $0$
$1$ $\frac{32}{221}$ $\frac{32}{221}$ $\frac{32}{221}$
$2$ $\frac{1}{221}$ $\frac{2}{221}$ $\frac{4}{221}$
    $\sum\text{xp}=\frac{34}{221}$ $\sum\text{x}^2\text{p}$
Mean $=\sum\text{xp}$ Mean $=\frac{34}{221}$ Variance $=\sum\text{x}^2\text{p}-(\text{Mean})^2$ $=\frac{36}{221}-\Big(\frac{34}{221}\Big)^2$ $=\frac{7956-1156}{48841}$ $=\frac{6800}{48841}$ Variance $=\frac{400}{2873}.$
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Question 925 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 935 Marks
Given the probability that A can solve a problem is $\frac{2}{3}$ and the probability that B can solve the same problem is $\frac{3}{5}$. Find the probability that none of the two will be able to solve the problem.
Answer
Given,
Probability that A can solve a problem $=\frac{2}{3}$
$\Rightarrow\ \text{P(A)}=\frac{2}{3}$
$=\text{P}(\overline{\text{A}})=1-\frac{2}{3}$
$\text{P}(\overline{\text{A}})=\frac{1}{3}$
Probability that B can solve the same problem $=\frac{3}{5}$
$\Rightarrow\ \text{P(B)}=\frac{3}{5}$
$\Rightarrow\ \text{P}(\overline{\text{B}})=1-\frac{3}{5}$
$\text{P}(\overline{\text{B}})=\frac{2}{5}$
P(None of them solve the problem)
$=\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$
$=\text{P}(\overline{\text{A}})\text{ P}(\overline{\text{B}})$
$=\frac{1}{2}\times\frac{2}{5}$
$=\frac{2}{15}$
Required probability $=\frac{2}{15}$
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Question 945 Marks
A bag contains 7 white, 5 black and 4 red balls. Four balls are drawn without replacement. Find the probability that at least three balls are black.
Answer
A bag contains 7 white, 5 black and 4 red balls.
Four balls are drawn without replacement
P (At least three balls are black)
= P(3 black balls and one not black or 4 black balls)
= P(3 black and one not black) + P(4 black balls)
$=\frac{{^5}\text{C}_3\times{^{11}}\text{C}_1}{{^{16}}\text{C}_4}+\frac{{^{5}}\text{C}_4}{^{16}\text{C}_4}$
$=\frac{\frac{5!}{3!2!}\times11+\frac{5!}{4!1!}}{\frac{16!}{4!12!}}\ \Big[\text{Since } ^\text{n}\text{C}_\text{r}=\frac{\text{n}!}{\text{r}!(\text{n}-\text{t})!}\Big]$
$=\frac{\frac{5.4}{2}\times11+5}{\frac{16\times15\times14\times13}{4\times3\times2}}$
$=\frac{(110+5)}{1820}$
$=\frac{115}{1820}$
$=\frac{23}{364}$
Required probability $=\frac{23}{364}$
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Question 955 Marks
An item is manufactured by three machines A, B and C. Out of the total number of items manufactured during a specified period, $50\%$ are manufactured on machine A, $30\%$ on B and $20\%$ on C, $2\%$ of the items produced on A and $2\%$ of items produced on B are defective and $3\%$ of these produced on C are defective. All the items stored at one godown. One item is drawn at random and is found to be defective. What is the probability that it was manufactured on machine A?
Answer
Consider the following events:
$E_1$ = Item is produced by machine A,
$E_2$ = Item is produced by machine B,
$E_3$ = Item is produced by machine C,
A = Item is defective
Clearly,
$\text{P}(\text{E}_1)=\frac{50}{100}=\frac{1}{2},\text{P}(\text{E}_2)=\frac{30}{100}=\frac{3}{10},\text{P}(\text{E}_3)=\frac{20}{100}=\frac{1}{5}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{2}{100},\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{2}{100},\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=\frac{3}{100}$
Required probability $=\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)$
$=\frac{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{ P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{1}{2}\times\frac{2}{100}}{\frac{1}{2}\times\frac{2}{100}+\frac{3}{10}\times\frac{2}{100}+\frac{1}{5}\times\frac{3}{100}}$
$=\frac{5}{11}$
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Question 965 Marks
In a hospital, there are 20 kidney dialysis machines and that the chance of any one of them to be out of service during a day is 0.02. Determine the probability that exactly 3 machines will be out of service on the same day.
Answer
let X = nimber of out of service machines
p = probability that machine will be out of service on the
same day $=\frac{2}{100}$
q = probability that machine will be in service on the
same day $=\frac{8}{100}$
$\text{P(X}=3)=$ probability exactly 3 machines will be out of service on the same day
$\text{P(X}=3)=\text{ }^{20}\text{C}_3\times\big(\frac{2}{100}\big)^3\big(\frac{8}{100}\big)^0$
$=1140\times0.000008=0.00912$
For low probability event Poisson' distribution is used instead of Binomial distribution. then,
$\lambda=\text{np}=20\times0.02=0.4$
$\text{P(X = r})=\frac{\text{e}^{-\lambda}\times\lambda^3}{\text{r}!}$
$\text{P(X}=3)=\frac{\text{e}^{-0.4}\times0.4^3}{3!}=0.6703\times\frac{0.064}{6}=0.0071$
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Question 975 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
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, $i. \ k ≥ 0,$
​​​​​​​$ \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 985 Marks
A box contains 100 tickets, each bearing one of the numbers from 1 to 100. If 5 tickets are drawn successively with replacement from the box, find the probability that all the tickets bear numbers divisible by 10.
Answer
Let P denote the probability of getting a ticket bearing number divisibal by 10, So
$\text{p}=\frac{10}{100}$ [Since there are 10, 20, 30, 40, 50, 60, 70, 80, 90, 100 which are divisible by 10]
$\text{p}=\frac{1}{10}$
$\text{q}=1-\frac{1}{10}$ [Since p + q = 1]
$\text{q}=\frac{9}{10}$
Let X denote the variable representing the number of tickets bearing a number divisible by 10 out of 5 tickets. probability of getting r tickets bearing a number divisible by 10 out ot n tickets is given by
$\text{P}(\text{X = r})=\text{ }^\text{n}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$=\text{ }^{\text{5}}\text{c}_{\text{r}}\big(\frac{1}{10}\big)^{\text{r}}\big(\frac{9}{10}\big)^{5-\text{r}}\dots(1)$
Probability of getting all the tickets bearing a number divisible by 10
$=\text{ }^5\text{c}_5\big(\frac{1}{10}\big)^5\big(\frac{9}{10}\big)^{5-5}$ [Using (1)]
$=1.\big(\frac{1}{10}\big)^5\big(\frac{9}{10}\big)^0$
$=\big(\frac{1}{10}\big)^5$
Required probability $=\big(\frac{1}{10}\big)^5$
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Question 995 Marks
In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins/loses.
Answer
When a die is thrown, probability of getting a six (p) $=\frac{1}{6}$
Therefore, $=\text{q}=1-\text{p}=1-\frac{1}{6}=\frac{5}{6}$
  1. If he gets a six in first throw, then,
Probability of getting a six $=\frac{1}{6}$
  1. If he does not get a six, in first throw, but he gets a six in the second throw, then
Its probability $=\frac{5}{6}\times \frac{1}{6}=\frac{5}{36}$

Probability that he does not get a six in first two throws and he gets a six in thied throw

$=\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}=\frac{25}{216}$

Probability that he does not get a six in any of the three throws $=\bigg(\frac{5}{6}\bigg)^3=\frac{125}{216}$

In first throw he gets a six, will receive Rs.1

If he gets a six in second throw, he will receive Rs.(1 - 1) = 0

If he gets a six in third throw, he will receive Rs.(- 1 - 1 + 1)=Rs. -1 =he will loss Rs. 1

Exapected value $=\frac{1}{6}\times1+\bigg(\frac{5}{6}\times\frac{1}{6}\bigg)\times0+\bigg(\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}\bigg)\times(-1)$

$=\frac{1}{6}-\frac{25}{216}=\frac{11}{216}\ (\text{loss})$
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Question 1005 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 first throw results in head,
B = the last throw results in tail.
Answer
A coin is tossed thrice
Samplw space = {HHT, HTT, THT, TTT, HHH, HTH, THH, TTH}
A = The first throw results in head
A = {HHT, HTH, HHH, HTT}
B = The last throw in tail
B = {HHT, HTT, THT, TTT}
$\text{A}\cap\text{B}=\{\text{HHT, HTT}\}$
$\text{P}(\text{A}\cap\text{B})=\frac{2}{8}=\frac{1}{4}$
$\text{P(A) }\text{P(B)}=\frac{1}{2},\frac{1}{2}$
$\text{P(A) }\text{P(B)}=\frac{1}{4}$
$\text{P(A)}.\text{P(B)}=\text{P}(\text{A}\cap\text{B})$
So, A and B are independent events.
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