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MATHS 3 Marks Question

Probability · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 183 questions.

Questions · Page 2 of 4

3 Marks Question

Question 513 Marks
From a pack of 52 cards, 4 are drawn one by one without replacement. Find the probability that all are aces (or kings).
Answer
A = First card Ace
B = Second card Ace
C = Third card Ace
D = Fourth card Ace
P (All four drawn are Ace, without replacement)
$=\text{P}(\text{A})\ \text{P}\Big(\frac{\text{B}}{\text{A}}\Big)\ \text{P}\Big(\frac{\text{C}}{\text{A}\cap\text{B}}\Big)\ \text{P}\Big(\frac{\text{D}}{\text{A}\cap\text{B}\cap\text{C}}\Big)$
$=\frac{4}{52}\times\frac{3}{51}\times\frac{2}{50}\times\frac{1}{49}$ [Since, there are four Ace in 52 cards]
$=\frac{1}{270725}$
Required Probabilty $=\frac{1}{270725}$
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Question 523 Marks
There are $6\%$ defective items in a large bulk of items. Find the probability that a sample of $8$ items will include not more than one defective item.
Answer
Let $x$ denote the number of defective items in a sampla of $8$ items. athen, $x$ follows a binomial distribution with $n = 8,$
$P = ($Probability of getting a defective item$) = 0.06$ and $q = 1 - P = 0.94$
$P(X = r) = ^8C_r(0.06)^{r }(0.94)^{8-r}, r = 0, 1, 2, 3, .....8$
The required probability = probebility of not more than one defective item
$=\text{P}(\text{X}\leq1)$
$=\text{P}(\text{X}=0)+\text{P}(\text{x}=1)$
$=\text{ }^8\text{C}_0 (0.06)^0(0.94)^{8-0}+\text{ }^8\text{C}_1(0.06)^1(0.94)^{8-1}$
$=(0.94)^8+8(0.06)(0.94)^7$
$=(0.94)^7\big\{0.94+0.48\big\}$
$=1.42(0.94)^7$
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Question 533 Marks
Can the mean of a binomial distribution be less than its variance$?$
Answer
Let $X$ be a binomial veriate with parameters $n$ and $p.$
Mean $= np$ varience $= npq$
Mean $-$ variance $= np - npq = np (1 - q) = np.p = np^2$ 
Mean $-$ variance $ >0$ Mean $ > $ variance
So, mean can never be less than variance.
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Question 543 Marks
In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is $\frac{5}{6}.$ What is the probability that he will knock down fewer than 2 hurdles?
Answer
p = probability of knocking down a hurdle $=1-\frac{5}{6}=\frac{1}{6}$
q =  probability of clearing a hurdle $=\frac{5}{6}$
n = 10
P(He will knock down fewer than 2 hurdles) $=\text{P}(0\leq{\text{x}\leq2})=\text{P}(\text{X}=0)+\text{P}(\text{X}=1)$
$=\text{C}(10, 0)\bigg(\frac{1}{6}\bigg)^0\bigg(\frac{5}{6}\bigg)^{10}+\text{C}(10, 1)\bigg(\frac{1}{6}\bigg)^1\bigg(\frac{5}{6}\bigg)^9$
$=\bigg(\frac{5}{6}\bigg)^9\bigg(\frac{5}{6}+\frac{10}{6}\bigg)=\frac{5}{2}\bigg(\frac{5}{6}\bigg)^9$
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Question 553 Marks
Two coins are tossed once. Find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ in each of the following:
A = No tail appears,
B = No head appears.
Answer
Sample space of two coins
{HH, HT, TH, TT}
A = No tail appears
A = {HH}
B = No head appears
B = {TT}
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{0}{1}$
$=0$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0$
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Question 563 Marks
Find the chance of drawing 2 white balls in succession from a bag containing 5 red and 7 white balls, the ball first drawn not being replaced.
Answer
Bag contains 5 red and 7 white balls
A = First ball white
B = Second ball white
P (2 white balls drawn without replacement)
$=\text{P}(\text{A})\text{ P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{7}{12}\times\frac{6}{11}$
$=\frac{7}{22}$
Required probability $=\frac{7}{22}$
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Question 573 Marks
A man wins a rupee for head and loses a rupee for tail when a coin is tossed. Suppose that he tosses once and quits if he wins but tries once more if he loses on the first toss. Find the probability distribution of the number of rupees the man wins.
Answer
Let $n$ denote the number of throws required to get a head and $X$ denote the amount won / lost.
He may get head on first toss or lose first and $2^{nd}$ toss or lose and won second toss probability distribution for $X$
$\text{Number of throws (n)}:$ $1$ $2$ $2$
$\text{Amount won/lost(x)}:$ $1$ $0$ $-2$
$\text{Probability P (X)}:$ $\frac{1}{2}$ $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$ $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$
So probability distribution is given by
$\text{X}$ $\text{P(X)}$
$0$ $\frac{1}{4}$
$1$ $\frac{1}{2}$
$-2$ $\frac{1}{4}$
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Question 583 Marks
Two coins are tossed once. Find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ in each of the following:
A = Tail appears on one coin,
B = One coin shows head.
Answer
Sample space of two coins
{HH, HT, TH, TT}
A = Tail appears on one coin
A = {HT, TH}
B = One coin shows head
B = {HT, TH}
$(\text{A}\cap\text{B})=\{{\text{HT}, \text{TH}}\}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{2}{2}$
Hence, $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=1$
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Question 593 Marks
A die is rolled. If the outcome is an odd number, what is the probability that it is prime?
Answer
A die is rolled.
A = A prime number on die
A = {2, 3, 5}
B = An odd number on die
B = {1, 3, 5}
$(\text{A}\cap\text{B})=\{3,5\}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}$
$=\frac{2}{3}$
Required probability $=\frac{2}{3}$
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Question 603 Marks
Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event 'the coin shows a tail', given that 'at least one die shows a 3'.
Answer
S = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6),(1, H), (2, H), (3, H), (4, H), (5, H), (1, T), (2, T), (3, T), (4, T), (5, T)}
$\therefore\ \text{n}(\text{S})=20$
P(first die shows a multiple of 3) $=\frac{12}{36}=\frac{1}{3}$
P(first die shows a number which is not a multiple of 3) $=\frac{4}{6}\times\frac{1}{2}+\frac{4}{6}\times\frac{1}{2}=\frac{8}{12}=\frac{2}{3}$
Let A = the coin shows a tail = {(1, T), (2, T), (4, T), (5, T)}
B = at least one die shows a 3 = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)}
$\text{A}\cap\text{B}=\phi$ $\text{n}(\text{A})=4,\ \ \ \ \text{n}(\text{B})=6,\ \ \ \ \text{n}(\text{A}\cap\text{B})=0$ $\text{P}(\text{B})=\frac{6}{36}=\frac{1}{6}\ \text{and}\ \text{P}(\text{A}\cap\text{B})=0$ $\text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{A}\ \cap\ \text{B})}{\text{P}(\text{B})}=\frac{0}{\frac{6}{36}}=0$
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Question 613 Marks
A speaks the truth 8 times out of 10 times. A die is tossed. He reports that it was 5. What is the probability that it was actually 5?
Answer
Let A denote the event that man reports that 5 occurs and E the event that 5 actually turns up.
$\therefore\text{P(E)}=\frac{1}{6}$ and $\text{P}(\overline{\text{E}})=1-\frac{1}{6}=\frac{5}{6}$
Also, $\text{P}\Big(\frac{\text{A}}{\text{E}}\Big)=$ Probability that man reports that 5 occurs given thet 5 actually turns up = Probability mab speaking the truth $=\frac{8}{10}=\frac{4}{5}$
$\text{P}\Big(\frac{\text{A}}{\overline{\text{E}}}\Big)=$ Probability thet man reports that 5 occurs given that 5 does not turns up = Probability not speaking the truth $=1-\frac{4}{5}=\frac{1}{5}$
$\therefore$ Required probability $\text{P}\Big(\frac{\text{E}}{\text{A}}\Big)=\frac{\text{P}(\text{E})\text{P}\Big(\frac{\text{A}}{\text{E}}\Big)}{\text{P}(\text{E})\text{P}\Big(\frac{\text{A}}{\text{E}}\Big)+\text{P}(\overline{\text{E}})\text{P}\Big(\frac{\text{A}}{\overline{\text{E}}}\Big)}$
$=\frac{\frac{1}{6}\times\frac{4}{5}}{\frac{1}{6}\times\frac{4}{5}+\frac{5}{6}\times\frac{1}{5}}=\frac{4}{9}$
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Question 623 Marks
Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find E(X).
Answer
S = {(1, 2), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1), (1, 3), (2, 3), (3, 2), (4, 2), (5, 2), (6, 2), (1, 4), (2, 4), (3, 4), (4, 3), (5, 3), (6, 3), (1, 5), (2, 5), (3, 5), (4, 5), (5, 4), (6, 4), (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6)} n(S) = 30 Let X denotes the larger of the two numbers obtained.
$\text{x}_i$ $\text{f}_i$ $\text{p}_i$ $\text{p}_i\text{x}_i$
$2$
$3$
$4$
$5$
$6$		 $2$
$4$
$6$
$8$
$10$		 $\frac{2}{30}$
$\frac{4}{30}$
$\frac{6}{30}$
$\frac{8}{30}$
$\frac{10}{30}$		 $\frac{4}{30}$
$\frac{12}{30}$
$\frac{24}{30}$
$\frac{40}{30}$
$\frac{60}{30}$
  $30$   $\sum\text{p}_i\text{x}_i=\frac{140}{30}$
$\text{E}(\text{X})=\sum\text{p}_i\text{x}_i=\frac{140}{30}=\frac{14}{3}=4\frac{2}{3}$
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Question 633 Marks
A random variable X has the following probability distribution:
Values of X: 0 1 2 3 4 5 6 7 8
P(X) a 3a 5a 7a 9a 11a 13a 15a 17a
Determine:
$\text{P}(\text{X}<3),\text{P}(\text{X}\geq3),\text{P}(0<\text{X}<5).$
Answer
$\text{P}(\text{X}<3)=\text{P}(0)+\text{P}(1)+\text{P}(2)$
$=\text{a}+3\text{a}+5\text{a}$
$=9\text{a}$
$=9\Big(\frac{1}{81}\Big)$
$\therefore\ \text{P}(\text{X}<3)=\frac{1}{9}$
$\text{P}(\text{X}\geq3)=1-\text{P}(\text{X}<3)=1-\frac{1}{9}=\frac{8}{9}$
$\text{P}(0<\text{X}<5)=\text{P}(1)+\text{P}(2)+\text{P}(3)+\text{P}(4)$
$=3\text{a}+5\text{a}+7\text{a}+9\text{a} $
$=24\text{a}$
$=24\Big(\frac{1}{81}\Big)$
$\therefore\ \text{P}(0<\text{X}<5)=\frac{8}{27}$
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Question 643 Marks
Three cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of spades. Hence, find the mean of the distribution.
Answer
We have,
p = probability of getting a spade in a draw $=\frac{13}{52}=\frac{1}{4}$ and $\text{q}=1-\text{p}=1-\frac{1}{4}=\frac{3}{4}$
Let X denote a success of getting a spade in a throw. Then,
X follows binomial distribution with parameters $\text{n}=3$ and $\text{p}=\frac{1}{4}$
$\therefore\text{P(X = r})=\text{ }^3\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{(3-\text{r})}=\text{ }^3\text{c}_{\text{r}}\big(\frac{1}{4}\big)^{\text{r}}\big(\frac{3}{4}\big)^{(3-\text{r})}=\frac{\text{ }^{3}\text{c}_{\text{r}}3^{(3-\text{r})}}{4^3}=\frac{27}{64}\Big(\frac{\text{ }^3\text{c}_{\text{r}}}{3^{\text{r}}}\Big),$ where, $\text{r}=0,1,2,3$
So, the probability distibution of X is given by:
$\text{P(X = r})=\frac{27}{64}\Big(\frac{\text{ }^3\text{C}_{\text{r}}}{3^{\text{r}}}\Big),$ where, $\text{r}=0,1,2,3$
Now,
$\text{Mean, E(X) = np}=3\times\frac{1}{4}=\frac{3}{4}$
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Question 653 Marks
If the probability distribution of a random variable of $X$ is given by
$X = x_i:$ $1$ $2$ $3$ $4$
$P(X = x_i):$ $2k$ $4k$ $3k$ $k$
Write tyhe value of $k.$
Answer
Here,
$X = x_i:$ $1$ $2$ $3$ $4$
$P(X = x_i):$ $2k$ $4k$ $3k$ $k$
Since, $\sum\text{P}(\text{X})=1$
$\Rightarrow\text{P}(1)+\text{P}(2)+\text{P}(3)+\text{P}(4)=1$
$\Rightarrow2\text{k}+4\text{k}+3\text{k}+\text{k}=1$
$\Rightarrow10\text{k}=1$
$\Rightarrow\text{k}=\frac{1}{10}$
$\Rightarrow\text{k}=0.1$
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Question 663 Marks
A speaks truth in 75% and B in 80% of the cases. In what percentage of cases are they likely to contradict each other in narrating the same incident?
Answer
Given
A speaks truth in 75% cases.
B speaks truth in 80% cases.
$\text{P(A)}=\frac{75}{100}\Rightarrow\text{P}(\overline{\text{A}})=\frac{25}{100}$
$\text{P(B)}=\frac{80}{100}\Rightarrow\text{P}(\overline{\text{B}})=\frac{20}{100}$
P(A and B contradict each other)
$=\text{P}\big[(\text{A}\cap\overline{\text{B}})\cup(\overline{\text{A}}\cap\text{B})\big]$
$=\text{P}(\text{A}\cap\overline{\text{B}}) +\text{P}(\overline{\text{A}}\cap\text{B})$
$=\text{P(A)}\text{ P}(\overline{\text{B}})+\text{P}(\overline{\text{A}})\text{ P(B)}$
$=\frac{75}{100}\times\frac{20}{100}+\frac{25}{100}\times\frac{80}{100}$
$=\frac{1500}{1000}+\frac{2000}{10000}$
$=\frac{3500}{10000}$
$=35\%$
Required probability $=35\%$
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Question 673 Marks
A laboratory blood test is $99\%$ effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for $0.5\%$ of the healthy person tested $($i.e. if a healthy person is tested, then, with probability $0.005,$ the test will imply he has the disease$).$ If $0.1$ percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?
Answer
Let $E_{1 }=$ The person selected is suffering from certain disease,
$E_{2 }=$ The person selected is not suffering from certain disease and $A =$ The doctor diagnoses correctly
$\text{Now}\ \ \text{P}(\text{E}_1)=0.1\%=\frac{1}{1000}=0.001,\ \text{P}(\text{E}_2)= 1-\frac{1}{1000}=\frac{999}{1000}=0.999,$
$\text{P}(\text{A}|\text{E}_1)=99\%=\frac{99}{100}=0.99\ \text{P}(\text{A}|\text{E}_2)=0.005\%$
Therefore, by Bayes’ theorem,
$\text{P}(\text{E}_1|\text{A})=\frac{\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)}{\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\text{P}(\text{A}|\text{E}_2)}$
$=\frac{0.01\times0.99}{.001\times0.99+0.999\times0.005}=\frac{990}{990+4995}=\frac{22}{133}$
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Question 683 Marks
A coin is tossed three times. Find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ in each of the following:
A = At most two tails,
B = At least one tail.
Answer
Sample space for three coins is given by
{HHH, HTH, THH, TTH, HHT, HTT, THT, TTT}
A = At most two tails
A = {HHH, HTH, THT, TTH, HHT, THT, HTT}
B = At least one tail
B = {HTH, THH, TTH, HHT, HTT, THT, TTT}
$(\text{A}\cap\text{B})=\{\text{HTH, THT, TTH, HHT, THT, HTT}\}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{6}{7}$
Hence, $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{6}{7}$
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Question 693 Marks
A pair of dice is thrown 6 times. If getting a total of 9 is considered a success, what is the probability of at least 5 successes?
Answer
Let X be the number of successes in 6 throws of the two dice. Probability of success = Probability of getting a total of 9 = Probability of getting (3, 6), (4, 5), (5, 4), (6, 3) out of 36 outcomes $\text{p}=\frac{4}{36}=\frac{1}{9},\text{q}=1-\text{p}=\frac{8}{9}\text{ and }\text{n}=6$ X follows a binomial distribution with $\text{n}=6,\text{p}=\frac{1}{9}$ and $\text{q}=\frac{8}{9}$$\text{P}(\text{X = r})=\text{ }^6\text{C}_{\text{r}}\big(\frac{1}{9}\big)^{\text{r}}\frac{8}{9}^{6-\text{r}}$
The required probability = Probability of at least 5 success $=\text{P}(\text{X}\geq5)$ $=\text{P}(\text{X}=5)+\text{P}(\text{X}=6)$ $=\text{ }^6\text{C}_56\big(\frac{1}{9}\big)^5\big(\frac{8}{9}\big)^{6-5}+\text{ }^6\text{C}_6\big(\frac{1}{9}\big)^6\big(\frac{8}{9}\big)^{6-6}$ $=\frac{6(8)+1}{9^6}=\frac{49}{9^6}$
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Question 703 Marks
Let $A$ and $B$ be two independent events such that $P(A) = p_1$ and $P(B) = p_2.$
Describe in words the events whose probabilities are:
$p_1 + p_2 = 2p_1p_2.$
Answer
As $p_1 + p_2 - 2p_1p_2 = (p_1 - p_1p_2) + (p_2 - p_1p_2)$
$= [P(A) - P(A) \times P(B)] + [P(B) - P(A) \times P(B)]$
And, $A$ and $B$ are indepepndet events.
i.e., $\text{P(A)}\times\text{P(B)}=\text{P}(\text{A}\cap\text{B})$
$\Rightarrow \ \text{p}_1+\text{p}_2-2\text{p}_1\text{p}_2=\big[\text{P(A)}-\text{P}(\text{A}\cap\text{B})\big]+\big[\text{P(B)}-\text{P}(\text{A}\cap\text{B})\big]$
$=\text{P(only A)}+\text{P(only B)}$
So, $P($only $A) + P($only $B) = p_1 + p_2 - 2p_1p_2$
Hence, $p_1 + p_2 - 2p_1p_2 = P($Exactly one of $A$ and $B$ occurs$).$
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Question 713 Marks
Find the probability that in 10 throws of a fair die, a score which is a multiple of 3 will be obtained in at least 8 of the throws.
Answer
Here success is a score which is multiple of 3 i.e. 3 or 6.
$\therefore\text{p}(3 \text{ or } 6)=\frac{2}{6}=\frac{1}{3}$
The probability of r success in 10 throws is given by
$\text{P(r)}=\text{ }^{10}\text{C}_{\text{r}}\big(\frac{1}{3}\big)^{\text{r}}\big(\frac{2}{3}\big)^{10-\text{r}}$
Now P(at least 8 success) =P(8) + P(9) + P(10)
$=\text{ }^{10}\text{C}_8\big(\frac{1}{3}\big)^8\big(\frac{2}{3}\big)^2+\text{ }^{10}\text{C}_9\big(\frac{1}{3}\big)^9\big(\frac{2}{3}\big)^1+\text{ }^{10}\text{C}_{10}\big(\frac{1}{3}\big)^{10}\big(\frac{2}{3}\big)^{0}$
$=\frac{1}{3^{10}}\big[45\times4+10\times2+1\big]$
$=\frac{20^1}{3^{10}}$
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Question 723 Marks
An urn contains four white and three red balls. Find the probability distribution of the number of red balls in three draws with replacement from the urn.
Answer
As three balls are drawn with replacement, the number of white balls, say X, follows binomial distribution with n = 3
$\text{p}=\frac{3}{7}$ and $\text{q}=\frac{4}{7}$
$\text{P}(\text{X = r})=\text{ }^{3}\text{C}_{\text{r}}\big(\frac{3}{7}\big)^{\text{r}}\big(\frac{4}{7}\big)^{3-\text{r}},\text{r}=0,1,2,3$
$\text{P}(\text{X}=0)=\text{ }^{3}\text{C}_0\big(\frac{3}{7}\big)^0\big(\frac{4}{7}\big)^{3-0}$
$\text{P}(\text{X}=1)=\text{ }^3\text{C}_1\big(\frac{3}{7}\big)^1\big(\frac{4}{7}\big)^{3-1}$
$\text{P}(\text{X}=2)=\text{ }^2\text{C}_2\big(\frac{3}{7}\big)^2\big(\frac{4}{7}\big)^{3-2}$
$\text{P}(\text{X}=3)=\text{ }^3\text{C}_3\big(\frac{3}{7}\big)^3\big(\frac{4}{7}\big)^{3-3}$
$\text{X}$ $0$ $1$ $2$ $3$
$\text{P(X)}$ $\frac{64}{343}$ $\frac{144}{343}$ $\frac{108}{343}$ $\frac{27}{343}$
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Question 733 Marks
Two cards are drawn successively without replacement from well shuffled pack of 52 cards. Find the probability distribution of the number of aces.
Answer
Two cards are drawn successively without replacement from a pack of 52 cards. Let X denote the number of aces drawn from pack out of 2 cards. Then, X can take the values 0, 1 and 2.
Now,
P(X = 0)
$=\frac{48}{52}\times\frac{47}{51}$
$=\frac{2256}{2652}$
$=\frac{188}{221}$
P(X = 1)
$=\frac{4}{52}\times\frac{48}{51}+\frac{48}{52}\times\frac{4}{51}$
$=\frac{384}{2652}$
$=\frac{32}{221}$
P(X = 2)
$=\frac{4}{52}\times\frac{3}{51}$
$=\frac{12}{2652}$
$=\frac{1}{221}$
So, required probability distribution is
$\text{X}:$
$0$
$1$
$2$
$\text{P}(\text{X}):$
$\frac{188}{221}$
$\frac{32}{221}$
$\frac{1}{221}$
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Question 743 Marks
A black and a red die are rolled.
Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
Answer
$\text{n}(\text{S})=6\times6=36$
Let A represents obtaining a sum greater than 9 and B represents black die resulted in a 5.
$\text{A}=\{(4, 6)\ (6, 4)\ (5, 5)\ (3, 6)\ (6, 3)\ (4, 5)\ (5, 4)\ (6, 5)\ (5, 6)\ (6, 6)\}\ \ \ \ \Rightarrow\ \ \ \ \text{n}(\text{A})=10$
$\text{P}\left(\text{A}\right)=\frac{\text{n}\left(\text{A}\right)}{\text{n}\left(\text{S}\right)}=\frac{10}{36}$
$\text{B}=\{(5, 1)\ (5, 2)\ (5, 3)\ (5, 4)\ (5, 5)\ (5, 6)\}\ \ \ \ \ \Rightarrow\ \ \ \ \text{n}(\text{B})=6$
$\text{P}\left(\text{B}\right)=\frac{\text{n}\left(\text{B}\right)}{\text{n}\left(\text{S}\right)}=\frac{6}{36}$
$ \text{A}\cap\text{B}=\left(55,\ 56\right)\ \ \ \ \ \ \Rightarrow\ \ \ \text{n}\left(\text{A}\cap\text{B}\right)=2$
$\text{P}\left(\text{A}\cap\text{B}\right)=\frac{2}{36}$
$\text{P}\left(\text{A}|\text{B}\right)=\frac{\text{P}\left(\text{A}\ \cap\ \text{B}\right)}{\text{P}\left(\text{B}\right)}=\frac{\frac{2}{36}}{\frac{6}{36}}=\frac{2}{6}=\frac{1}{3}$
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Question 753 Marks
From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.
Answer
n(S) = 30, A = {6 defective bulbs} ⇒ n(A) = 6
$\text{P}=\frac{\text{n}(\text{A})}{\text{n}(\text{S})}=\frac{6}{30}=\frac{1}{5}\ \text{and}\ \text{q}=1-\frac{1}{5}=\frac{4}{5}$
n = 4 (4 bulbs are drawn with replacement), r = 0, 1, 2, 3, 4
$\text{P}(\text{X}=0)=(\text{q})^4=\bigg(\frac{4}{5}\bigg)^4=\frac{256}{625}$
$\text{P}(\text{X}=1)=\ ^4\text{C}_1(\text{p})\text{q}^3=4\times\frac{1}{5}\times\bigg(\frac{4}{5}\bigg)^3=\frac{256}{625}$
$\text{P}(\text{X}=2)=\ ^4\text{C}_2\text{p}^2\text{q}^2=6\times\bigg(\frac{1}{5}\bigg)^2\times\bigg(\frac{4}{5}\bigg)^2=\frac{96}{625}$
$\text{P}(\text{X}=3)=\ ^4\text{C}_3\text{p}^3\text{q}=4\times\bigg(\frac{1}{5}\bigg)^3\times\bigg(\frac{4}{5}\bigg)=\frac{16}{625}$
$\text{P}(\text{X}=4)=\ ^4\text{C}_4\text{p}^4=\times\bigg(\frac{1}{5}\bigg)^4=\frac{1}{625}$
Probability distribution:
$\text{X}_i$ $0$ $1$ $2$ $3$ $4$
$\text{P}(\text{X}_i)$ $\frac{256}{625}$ $\frac{256}{625}$ $\frac{96}{625}$ $\frac{16}{625}$ $\frac{1}{625}$
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Question 763 Marks
A manufacturer has three machine operators $A, B$ and $C.$ The first operator A produces $1\%$ defective items, where as the other two operators $B$ and $C$ produce $5\%$ and $7\%$ defective items respectively. $A$ is on the job for $50\%$ of the time, $B$ is on the job for $30\%$ of the time and $C$ is on the job for $20\%$ of the time. A defective item is produced, what is the probability that it was produced by $A?$
Answer
Let $E_{1 }=$ the item is manufactured by the operator $A,$
$E_{2 }=$ the item is manufactured by the operator $B,$
$E_{3 }=$ the item is manufactured by the operator $C$ and
$A =$ the item is defective
$\text{Now}\ \ \text{P}(\text{E}_1)=\frac{50}{100},\ \text{P}(\text{E}_2)=\frac{30}{100},\ \text{P}(\text{E}_3)=\frac{20}{100}$
Now $\text{P}(\text{A}|\text{E}_1) = P($item drawn is manufactured by operator $A) =\frac{1}{100}$
Similarly, $\text{P}(\text{A}|\text{E}_2)=\frac{5}{100}\ \text{and}\ \text{P}(\text{A}|\text{E}_3)=\frac{7}{100}$
Now Required probability $=$ Probability that the item is manufactured by operator A given that the item drawn is defective
$\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{P}(\text{A}|\text{E}_3)}$
$ =\frac{\frac{50}{100}\times\frac{1}{100}}{\frac{50}{100}\times\frac{1}{100}+\frac{30}{100}\times\frac{5}{100}+\frac{20}{100}\times\frac{7}{100}}=\frac{50}{50+150+40}=\frac{5}{34}$
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Question 773 Marks
A laboratory blood test is $99\%$ effective in detecting a certain disease when its infection is present. However, the test also yields a false positive result for $0.5\%$ of the healthy person tested $($i.e. if a healthy person is tested, then, with probability $0.005,$ the test will imply he has the disease$).$ If $0.1\%$ of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?
Answer
Let $E_1$ and $E_2$ denote the events that a person has a disease and a person has no disease, respectively.
$E_1$ and $E_2$ are complimentary to each other.
$\therefore P(E_1) + P(E_2) = 1$
$\Rightarrow P(E_2) = 1 - P(E_1) = 1 - 0.001 = 0999$
let A denote the event that the blood test result is positive.
$\therefore P(E_1) = 0.1\% = 0.001$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=90\%=0.99$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\big)=0.5\%=0.005$
Using Baye's 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)}$
$=\frac{0.001\times0.99}{0.001\times0.99+0.999\times0.005}$
$=\frac{990}{5985}=\frac{22}{133}$
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Question 783 Marks
An urn contain 5 red and 2 black balls. Two balls rendomly selected. Let X represent the number of black ball. What are the possible values of X. Is X a random variable?
Answer
Urn has 5 red and 2 black balls. 2 balls are rendomly selected. Here, X denote the numbers of black balls. So, possible values of X = 0, 1, 2 $\text{P}(\text{x}=0)=\text{P}\big(\overline{\text{B}}_1\big)\times\text{P}\big(\overline{\text{B}}_2\big)$ $=\frac{5}{7}\times\frac{5}{7}$ $=\frac{25}{49}$ $\text{P}(\text{X}=1)=\text{P}(\text{B}_1)\text{P}\big(\overline{\text{B}}_2\big)+\text{P}\big(\overline{\text{B}}_1\big)\text{P}(\text{B}_2)$ $=\frac{2}{7}\times\frac{5}{7}+\frac{5}{7}\times\frac{2}{7}$ $=\frac{20}{49}$ $\text{P}(\text{X}=2)=\text{P}(\text{B}_1)\text{P}(\text{B}_2)$ $=\frac{2}{7}\times\frac{2}{7}$ $=\frac{4}{49}$Now,
$\text{P}(\text{X}=0)+\text{P}(\text{X}=1)+\text{P}(\text{X}=2)$ $=\frac{25}{49}+\frac{20}{49}+\frac{4}{49}$ $=\frac{49}{49}$ $=1$ So, $\sum\text{P}(\text{X})=1$ Therefore, X is a random variable.
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Question 793 Marks
The probability that at least one of the two events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.3, evaluate $\text{P}\bar{(\text{A})}+\text{P}\bar{(\text{B})}.$
Answer
We know that $\text{A}\cup\text{B}$ denotes the occurrence of at least one of A and B and $\text{A}\cap\text{B}$ denotes the occurrence of both A and B, simultaneously.
Thus, $\text{P}(\text{A}\cup\text{B})=0.6$ and $\text{P}(\text{A}\cap\text{B})=0.3$
Also, $\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow0.6=\text{P}(\text{A})+\text{P}(\text{B})-0.3$
$\Rightarrow\text{P}(\text{A})+\text{P}(\text{B})=0.9$
$\Rightarrow\Big[1-\text{P}\bar{(\text{A})}\Big]+\Big[1-\text{P}\bar{(\text{B})}\Big]=0.9$ $\Big[\because\text{P}(\text{A})=1-\text{P}\bar{(\text{A})}\text{ and }\text{P}(\text{B})=1-\text{P}\bar{(\text{B})}\Big]$
$\Rightarrow\text{P}\bar{(\text{A})}+\text{P}\bar{(\text{B})}=2-0.9=1. 1$
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Question 803 Marks
If the mean of a binomial distribution is 20 and its standard deviation is 4, find p.
Answer
let n and p be the parameter of binomial distribution. So
Given,
$\text{Mean = np}=20\dots(1)$
$\text{Standard deviation } =\sqrt{\text{npq}}=4$
Squaring both the sides,
$\text{npq}=16\dots(2)$
Dividing equation (2) by (1),
$\frac{\text{npq}}{\text{np}}=\frac{16}{20}$
$\text{q}=\frac{4}{5}$
$\text{p}=1-\text{q}$ [Since p + q = 1]
$=1-\frac{4}{5}$
$\text{p}=\frac{1}{5}$
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Question 813 Marks
Five dice are thrown simultaneously. If the occurrence of 3, 4 or 5 in a single die is considered a success, find the probability of at least 3 successes.
Answer
Let X denote the occurrence of 3, 4 or 5 in a single die. Then, X follows binomial distribution with n = 5.
Let p = probability of getting 3, 4 or 5 in a single die
$\text{p}=\frac{3}{6}=\frac{1}{2}$
$\text{q}=1-\frac{1}{2}=\frac{1}{2}$
$\text{P}(\text{X = r})=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}2{}\big)^{5-\text{r}}$
P(at least 3 successes) $=\text{P}(\text{X}\geq3)$
$=\text{P(X = 3)}+\text{P}(\text{X}=4)+\text{P}(\text{X}=5)$
$=\text{ }^5\text{C}_3\big(\frac{1}{2}\big)^3\big(\frac{1}{2}\big)^{5-3}+\text{ }^5\text{C}_4\big(\frac{1}{2}\big)^4\big(\frac{1}{2}\big)^{5-4}+\text{ }^5\text{C}_5\big(\frac{1}{2}\big)^5\big(\frac{1}{2}\big)^{5-5}$
$=\frac{\text{ }^5\text{C}_3+\text{ }^5\text{C}_4+\text{ }^5\text{C}_5}{2^5}$
$=\frac{1}{2}$
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Question 823 Marks
A coin is tossed 5 times. If X is the number of heads observed, find the probability distribution of X.
Answer
Let X = number of heads in 5 tosses. Then the binomial distribution
for X has $\text{n}=5,\text{p}=\frac{1}{2}$ and $\text{q}=\frac{1}{2}$
$\text{P(X = r)}=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{5-\text{r}},\text{r}=0, 1, 2, 3, 4, 5$
$=\frac{\text{ }^5\text{C}_{\text{r}}}{2^5}$
Substituting r = 0, 1, 2, 3, 4, 5 we get the following probability disrtribution.
$\text{X}$ $0$ $1$ $2$ $3$ $4$ $5$
$\text{P(X)}$ $\frac{1}{32}$ $\frac{5}{32}$ $\frac{10}{32}$ $\frac{10}{32}$ $\frac{5}{32}$ $\frac{1}{32}$
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Question 833 Marks
A fair die is tossed twice. If the number appearing on the top is less than 3, it is success. Find the probability distribution of number of successes.
Answer
Let X denote the event of getting number less than 3 (1 or 2) on throwing the die. Then, X can take the values 0, 1 and 2.
Now,
$\text{P}(\text{X}=0)=\frac{16}{36}=\frac{4}{9}$
$\text{P}(\text{X}=1)=\frac{16}{36}=\frac{4}{9}$
$\text{P}(\text{X}=2)=\frac{4}{36}=\frac{1}{9}$
Thus, the probability distribution of X is given by
$\text{X}$
$0$
$1$
$2$
$\text{P}(\text{X})$
$\frac{4}{9}$
$\frac{4}{9}$
$\frac{1}{9}$
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Question 843 Marks
In a 20-question true-false examination, suppose a student tosses a fair coin to determine his answer to each question. For every head, he answers 'true' and for every tail, he answers 'false'. Find the probability that he answers at least 12 questions correctly.
Answer
Let p be the probability of answering a true. So
$\text{p}=\frac{1}{2}$
$\text{q}=1-\frac{1}{2}$ [Since p +q = 1]
$=\frac{1}{2}$
Thus the probability that he answers at least 12 questions correctly among 20 questions is
$\text{P(X}\geq12)=\text{P(X}=12)+\text{P(X}=13)+\text{P(X}=14)+\text{P(X}=15)\\+\text{P(X}=16)+\text{P(X}=17)+\text{P(X}=18)+\text{P(X}=19)+\text{P(X}=20)$
$$$=\big(\frac{1}{2}\big)^{20}\big\{\text{ }^{20}\text{C}_{12}+\text{ }^{20}\text{C}_{13}+\text{ }^{20}\text{C}_{14}+\text{ }^{20}\text{C}_{15}\\+\text{ }^{20}\text{C}_{16}+\text{ }^{20}\text{C}_{17}+\text{ }^{20}\text{C}_{18}+\text{ }^{20}\text{C}_{19}+\text{ }^{20}\text{C}_{20}\big\}$
$\frac{\text{ }^{20}\text{C}_{12}+\text{ }^{20}\text{C}_{13}+\text{ }^{20}\text{C}_{14}+\text{ }^{20}\text{C}_{15}+\text{ }^{20}\text{C}_{16}+\text{ }^{20}\text{C}_{17}+\text{ }^{20}\text{C}_{18}+\text{ }^{20}\text{C}_{19}+\text{ }^{20}\text{C}_{20}}{2^{20}}$
Therefore, the required answer is
$\frac{\text{ }^{20}\text{C}_{12}+\text{ }^{20}\text{C}_{13}+\text{ }^{20}\text{C}_{14}+\text{ }^{20}\text{C}_{15}+\text{ }^{20}\text{C}_{16}+\text{ }^{20}\text{C}_{17}+\text{ }^{20}\text{C}_{18}+\text{ }^{20}\text{C}_{19}+\text{ }^{20}\text{C}_{20}}{2^{20}}$
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Question 853 Marks
Of the students in a college, it is known that $60\%$ reside in a hostel and $40\%$ do not reside in hostel. Previous year results report that $30\%$ of students residing in hostel attain A grade and $20\%$ of ones not residing in hostel attain A grade in their annual examination. At the end of the year, one students is chosen at random from the college and he has an A grade. What is the probability that the selected student is a hosteler?
Answer
Let $A, E_1$ and $E_2$ denote the events that the selected student attains grade $A,$ resedes in a hostel and does not reside in a hostel, respectively.
$\therefore\ \text{P}(\text{E}_1)=\frac{60}{100}$
$\text{P}(\text{E}_1)=\frac{40}{100}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{30}{100}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{20}{100}$
Using Baye's theorem,
Required probability $\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{60}{100}\times\frac{30}{100}}{\frac{60}{100}\times\frac{30}{100}+\frac{40}{100}\times\frac{20}{100}}$
$=\frac{18}{18+8}=\frac{9}{13}$
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Question 863 Marks
Assume that the chances of a patient having heart attack is $40\%.$ It is also assumed that meditation and yoga course reduces 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 and patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?
Answer
Let $A, E_1,$ and $E_2$ respectively denote the events that a person has heart attack, the selected person followed the course of yoga and meditation, and the person adopted the drug prescription.
$\therefore\ \text{P(A)}=0.40$
$\text{P}(\text{E}_1)=\text{P}(\text{E}_2)=\frac{1}{2}$
$\text{P}(\text{A}|\text{E}_1)=04.\times0.70=0.28$
$\text{P}(\text{A}|\text{E}_2)=0.40\times0.75=0.30$
Probability that the patient suffering a heart attack followe a course of meditation and yoga is given by $P(E_1|A).$
$\text{P}(\text{E}_1|\text{A})=\frac{\text{P}(\text{E}_1)\times\text{P}(\text{A}|\text{E}_1)}{\text{P}(\text{E}_1)\times\text{P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\times\text{P}(\text{A}|\text{E}_2)}$
$=\frac{\frac{1}{2}\times0.28}{\frac{1}{2}\times0.28+\frac{1}{2}\times0.30}$
$=\frac{14}{29}$
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Question 873 Marks
Two cards are drawn from a well shuffled pack of 52 cards. Find the probability distribution of the number of aces.
Answer
Let X denote number of aces in a sample of 2 cards drawn.
There are four aces in a pack of 52 cards.
So, X can have values 0, 1, 2
Now,
P(X = 0)
$=\frac{\text{ }^{48}\text{C}_2}{\text{}^{52}\text{C}_2}$
$=\frac{2256}{2652}$
$=\frac{188}{221}$
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,
$\text{X}:$
$0$
$1$
$2$
$\text{P}(\text{X}):$
$\frac{188}{221}$
$\frac{32}{221}$
$\frac{1}{221}$
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Question 883 Marks
Assume that on an average one telephone number out of 15 called between 2 P.M. and 3 P.M. on week days is busy. What is the probability that if six randomly selected telephone numbers are called, at least 3 of them will be busy?
Answer
Let X be the number of busy calls for 6 randomly selected telephone nimbers.
X following a binomial distribution with $\text{n}=6;\text{p}=1$ out of $15=\frac{1}{15}$ and $\text{q}=\frac{14}{15}$
$\text{p}(\text{X = r})=\text{ }^6\text{C}_{\text{r}}\big(\frac{1}{15}\big)^{\text{r}}\big(\frac{14}{15}\big)^{6-\text{r}}$
Probability that at least 3 of them are busy $=\text{P}(\text{X}\geq3)$
$=1-\big\{\text{P}(\text{X}=0)+\text{P}(\text{X}=1)+\text{P}(\text{X}=2)\big\}$
$=1-\big\{\text{ }^6\text{C}_0\big(\frac{1}{15}\big)^0\big(\frac{14}{15}\big)^{6-0}+\text{ }^6\text{C}_1\big(\frac{1}{15}\big)^1\big(\frac{14}{15}\big)^{6-1}+\text{ }^6\text{C}_2\big(\frac{1}{15}\big)^2\big(\frac{14}{15}\big)^{6-2}\big\}$
$=1-\big\{\big(\frac{14}{15}\big)^6+\frac{6}{15}\big(\frac{14}{15}\big)^5+\frac{1}{15}\big(\frac{14}{15}\big)^4\big\}$
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Question 893 Marks
From a pack of 52 cards, two are drawn one by one without replacement. Find the probability that both of them are kings.
Answer
A = first card is king
B = Second card is also king
Probability of getting two kings (Without replacement)
$=\text{P(A) }\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{4}{52}\times\frac{3}{51}$ [Since, 4 kings out of 52 cards.]
$=\frac{1}{13}\times\frac{1}{17}$
$=\frac{1}{221}$
Required probability $=\frac{1}{221}$
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Question 903 Marks
A bag contains $6$ red and $8$ black balls and another bag contains $8$ red and $6$ black balls. A ball is drawn from the first bag and without noticing its colour is put in the second bag. A ball is drawn from the second bag. Find the probability that the ball drawn is red in colour.
Answer
Given,
Bag $(1)$ contains $6$ red $(R_1)$ and $8$ black $(B_1)$ balls
Bag $(2)$ contains $8$ red $(R_2)$ and $6$ black $(B_2)$ balls
A ball is drawn from hte first bag and without noticing is colour is pur in the bag $(2).$
Then a ball is drawn from second bag and it is red.
$P($One red ball from bag $2)$
$=\text{P}\big((\text{B}_1\cap\text{P}_2)\cup(\text{R}_1\cap\text{R}_2)\big)$
$=\text{P}(\text{B}_1\cap\text{P}_2)+\text{P}(\text{R}_1\cap\text{R}_2)$
$=\text{P}(\text{B}_1)\text{ P}\Big(\frac{\text{R}_2}{\text{B}_1}\Big)+\text{P}(\text{R}_1)\text{ P}\Big(\frac{\text{R}_2}{\text{R}_1}\Big)$
$=\frac{8}{14}\times\frac{8}{15}+\frac{6}{14}\times\frac{9}{15}$
$=\frac{64+54}{210}$
$=\frac{118}{210}$
$=\frac{59}{105}$
Required probability $=\frac{59}{105}$
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Question 913 Marks
Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among 100 students, what is the probability that:
You both enter the same section?
Answer
Out of 100 students two friends can enter sections in 100C2 ways.
Let
A = Event both enter in section A (40 students)
B = Event both enter in section B (60 students)
$\text{P(A)}=\frac{^{40}\text{C}_2}{^{100}\text{C}_2},\text{P(B)}=\frac{^{60}\text{C}_2}{^{100}\text{C}_2}$
$=\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}$
$=\frac{^{40}\text{C}_2+^{60}\text{C}_2}{^{100}\text{C}_2}$
$=\frac{\frac{40\times39}{2}+\frac{60+59}{2}}{\frac{100\times99}{2}}$
$=\frac{780+1770}{4950}$
$=\frac{2550}{4950}$
$=\frac{17}{33}$
P (Both enter same section) $=\frac{17}{33}$
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Question 923 Marks
In answering a question on a multiple choice test a student either knows the answer or guesses. Let $\frac{3}{4}$ be the probability that he knows the answer and $\frac{1}{4}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\frac{1}{4}$. What is the probability that a student knows the answer given that he answered it correctly$?$
Answer
Let $A, E_1$ and $E_2$ denote the events that the answer is correct, the student knows the answer and the student guesses the answer, respectively.
$\therefore\ \text{P}(\text{E}_1)=\frac{3}{4}$
$\text{P}(\text{E}_2)=\frac{1}{4}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=1$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{1}{4}$
Using Baye's theorem, we get
Required probability $\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{3}{4}\times1}{\frac{3}{4}\times1+\frac{1}{4}\times\frac{1}{4}}$
$=\frac{3}{3+\frac{1}{4}}=\frac{12}{13}$
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Question 933 Marks
A bag contains 8 red and 6 green balls. Three balls are drawn one after another without replacement. Find the probability that at least two balls drawn are green.
Answer
A bag contains 8 red and 6 green balls.
Three balls are drawn without replacement
P (At least 2 balls are green)
$=\text{P}\big[(\text{G}_1\cap\text{G}_2\cap\text{R}_1)\cup(\text{G}_1\cap\text{R}_1\cap\text{G}_2)\\\cup(\text{R}_1\cap\text{G}_1\cap\text{G}_2)\cup(\text{G}_1\cap\text{G}_2\cap\text{G}_3)\big]$
$=\text{P}\big(\text{G}_1\cap\text{G}_2\cap\text{R}_1)+\text{ P}(\text{G}_1\cap\text{R}_1\cap\text{G}_2)\\+\text{ P}(\text{R}_1\cap\text{G}_1\cap\text{G}_2)+\text{ P}(\text{G}_1\cap\text{G}_2\cap\text{G}_3)$
$=\text{P}\big(\text{G}_1)\text{ P}\Big(\frac{\text{G}_2}{\text{G}_1}\Big)\text{P}\Big(\frac{\text{R}_1}{\text{G}_1\cap\text{G}_2}\Big)\\+\text{P}(\text{G}_1)\text{ P}\Big(\frac{\text{G}_2}{\text{G}_1}\Big)\text{ P}\Big(\frac{\text{G}_3}{\text{G}_1\cap\text{G}_2}\Big)$
$=\frac{6}{14}\times\frac{5}{13}\times\frac{8}{12}+\frac{6}{14}\times\frac{8}{13}\times\frac{5}{12}\\+\frac{8}{14}\times\frac{6}{13}\times\frac{5}{12}+\frac{6}{14}\times\frac{5}{13}\times\frac{4}{12}$
$=\frac{1}{14}\times\frac{1}{13}\times\frac{1}{12}\times(240+240+240+120)$
$=\frac{840}{14\times13\times12}$
$=\frac{5}{13}$
Required probability $=\frac{5}{13}$
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Question 943 Marks
Ten cards numbered 1 through 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?
Answer
Sample space, S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Consider the given events.
A = An even number on the card
B = A number more than 3 on the card
Clearly,
A = {2, 4, 6, 8, 10}
B = {4, 5, 6, 7, 8, 9, 10}
Now,
$\text{A}\cap\text{B}=\{4, 6, 8, 10\}$
$\therefore \text{Required probability} =\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}=\frac{4}{7}$
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Question 953 Marks
Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are possible values of X?
Answer
For one coin, $\text{S}=\left\{\text{H, T}\right\}\ \ \Rightarrow\ \ \text{n}(\text{S})=2$
Let A represents head $\Rightarrow\ \ \text{n}(\text{A})=1$
$\therefore\ \ \text{P}(\text{A})=\frac{1}{2}\ \text{and}\ \text{P}(\overline{\text{A}})=\frac{1}{2}$
n = 6, r = 0, 1, 2, 3, 4, 5, 6
$\text{P}(\text{X=0})=\Big[\text{P}(\overline{\text{X}})\Big]^6=\bigg(\frac{1}{2}\bigg)^6=\frac{1}{64}$
$\text{P}(\text{X}=1)=6\ \text{P}(\text{X})\Big[\text{P}(\overline{\text{X}})\Big]^6=6\times\Big(\frac{1}{2}\Big)^6=\frac{6}{64}$
$\text{P}(\text{X}=2)=15[\text{P}(\text{X})]^2\Big[\text{P}(\overline{\text{X}})\Big]^4=15\times\bigg(\frac{1}{2}\bigg)^6=\frac{15}{64}$
$\text{P}(\text{X}=3)=20[\text{P}(\text{X})]^3\Big[\text{P}(\overline{\text{X}})\Big]^3=20\times\bigg(\frac{1}{2}\bigg)^6=\frac{20}{64}$
$ \text{P}(\text{X}=4)=15[\text{P}(\text{X})]^4\Big[\text{P}(\overline{\text{X}})\Big]^2=15\times\bigg(\frac{1}{2}\bigg)^6=\frac{15}{64}$
$\text{P}(\text{X}=5)=6\ [\text{P}(\text{X})]^5\Big[\text{P}(\overline{\text{X}})\Big]=6\times\bigg(\frac{1}{2}\bigg)^6=\frac{6}{64}$
$\text{P}(\text{X}=6)=\Big[\text{P}(\overline{\text{X}})\Big]^6=\bigg(\frac{1}{2}\bigg)^6=\frac{1}{64}$
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Question 963 Marks
If for a binomial distribution P(X = 1) = P(X = 2) = α, write P(X = 4) in terma of α.
Answer
For binomial distribution of X, $\text{P(X = r})=\text{ }^{\text{n}}\text{C}_{\text{r}}(\text{p})^{\text{r}}(\text{q})^{\text{n}-\text{r}},\text{r}=0,1,2,\dots,\text{n}$ $\text{P(X}=1)=\text{np(q)}^{\text{n}-1}$ $\text{P(X}=2)=\text{ }^{\text{n}}\text{C}_2\text{p}^2(\text{q})^{\text{n}-2}$ $\Rightarrow\text{np(q)}^{\text{n}-1}=\text{ }^{\text{n}}\text{C}_2\text{p}^2(\text{q})^{\text{n}-2}=\alpha$ Simplifying the above equation we get, $\text{q}=\frac{\text{n}-1}{2}\text{p}$ $\Rightarrow2\text{q}=\text{np}-\text{p}$ On putting, $\text{q}=1-\text{p}$ we get $2-2\text{p}=\text{np}-\text{p}$ $\text{p(n+1})=2\dots(1)$ Also, $\text{P(X}=1)=\alpha$ $\Rightarrow\text{np}(1-\text{p})^{\text{n}-1}=\alpha\dots(2)$ Note: We cannot find the value of n as (1) and (2) are not linear and hence we cannot find the value of P(X = 4)Alternate Answer
Binomial distribution $\text{ }^{\text{n}}\text{C}_{\text{x}}\text{p}^{\text{x}}\text{q}^{(\text{n}-\text{x})}$ let X be the discrete variable, n the sample size $\text{P(X}=1)=\text{ }^{\text{n}}\text{C}_1\text{p}^1\text{q}^{(\text{n}-1)}$ $\text{P(X}=2)=\text{ }^{\text{n}}\text{C}_2\text{p}^2\text{q}^{(\text{n}-2)}$ Given $\text{P(X}=1)=\text{P(X}=2)=\alpha$ $\text{ }^{\text{n}}\text{C}_1\text{p}^1\text{q}^{(\text{n}-1)}=\text{ }^{\text{n}}\text{C}_2\text{p}^2\text{q}^{(\text{n}-2)}=\alpha$ $\text{npq}^{(\text{n}-1)}=\alpha\Rightarrow\text{q}^{\text{n}}=\frac{\alpha}{\text{n}}\times\frac{\text{q}}{\text{p}}$ $\text{P(X}=4)=\text{ }^{\text{n}}\text{C}_4\text{p}^4\text{q}^{(\text{n}-4)}=\text{ }^{\text{n}}\text{C}_4\text{p}^4\frac{\text{q}^{\text{n}}}{\text{q}^4}$ $=\text{ }^{\text{n}}\text{C}_4\text{p}^4\frac{1}{\text{q}^4}\times\frac{\alpha}{\text{n}}\times\frac{\text{q}}{\text{p}}$ $=\text{ }^{\text{n}}\text{C}_4\alpha\times\Big(\frac{\text{p}}{\text{q}}\Big)^3$
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Question 973 Marks
It is known that 10% of certain articles manufactured are defective. What is the probability that in a random sample of 12 such articles, 9 are defective?
Answer
The repeated selections of articles in a random sample space are Bernoulli trails. Let X denote the number of times of selecting defective articles in a random sample space of 12 articles.
Clearly, X has a binomial distribution with n = 12 and p = 10% $=\frac{10}{100}=\frac{1}{10}$
$\therefore\ \text{q}=1-\text{p}=1-\frac{1}{10}=\frac{9}{10}$
$\therefore\ \text{P}(\text{X=x})=\ =^\text{n}\text{C}_\text{x}\text{q}^\text{n-x}\text{p}^\text{x}=\ ^{12}\text{C}_\text{x}\bigg(\frac{9}{10}\bigg)^{12-\text{x}}\cdot\bigg(\frac{1}{10}\bigg)^\text{x}$
P(selecting 9 defective articles) $=\ ^{12}\text{C}_{9}\bigg(\frac{9}{10}\bigg)^3\bigg(\frac{1}{10}\bigg)^{9}$
$=220\cdot\frac{9^3}{10^3}\cdot\frac{1}{10^9}$
$=\frac{22\times9^3}{10^{11}}$
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Question 983 Marks
A bag contains 4 red and 6 black balls. Three balls are drawn at random. Find the probability distribution of the number of red balls.
Answer
A bag contains 4 red and 6 black balls. Three balls are drawn.
Let X denote number of red balls out of three drawn.
Then X = 0, 1, 2, 3.
So,
P(no red balls) = P(X = 0)
$=\frac{\text{}^6\text{C}_3}{\text{}^{10}\text{C}_3}$
$=\frac{20}{120}$
$=\frac{1}{6}$
P(one red balls) = P(X = 1)
$=\frac{\text{}^4\text{C}_1\times\text{}^6\text{C}_2}{\text{}^{10}\text{C}_3}$
$=\frac{60}{120}$
$=\frac{1}{2}$
P(two red balls) = P(X = 2)
$=\frac{\text{}^4\text{C}_2\times\text{}^6\text{C}_1}{\text{}^{10}\text{C}_3}$
$=\frac{36}{120}$
$=\frac{3}{10}$
Pall three red) = P(X = 3)
$=\frac{\text{}^4\text{C}_3}{\text{}^{10}\text{C}_3}$
$=\frac{4}{120}$
$=\frac{1}{30}$
The required probability distribution is
$\text{X}:$
$0$
$1$
$2$
$3$
$\text{P}(\text{X}):$
$\frac{1}{6}$
$\frac{1}{2}$
$\frac{3}{10}$
$\frac{1}{30}$
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Question 993 Marks
A fair die is rolled. Consider events E = $\{1,\ 3,\ 5\},\ \text{F}=\{2,\ 3\}\ \text{and}\ \text{G}=\{2,\ 3,\ 4,\ 5\}.\ \text{Find}:$ $\text{P}\Big[(\text{E}\cup\text{F})|\text{G}\Big]\ \text{and}\ \text{P}\Big[(\text{E}\cap\text{F})|\text{G}\Big]$ 
Answer
$ \text{P}\left(\text{G}\right)=\frac{\text{n}\left(\text{G}\right)}{\text{n}\left(\text{S}\right)}=\frac{4}{6}$
$\text{E}\cup\text{F}=\left(1,\ 2,\ 3,\ 5\right)\ \ \ \ \text{and}\ \text{G}\ \left(2,\ 3,\ 4,\ 5\right)$
$\left(\text{E}\cup\text{F}\right)\cap\text{G}=\left(2,\ 3,\ 5\right)\ \Rightarrow\ \ \ \ \text{n}\left[\left(\text{E}\cup\text{F}\right)\cap\text{G}\right]=3$
$\text{P}\left[\left(\text{E}\ \cup\ \text{F}\right)\cap\text{G}\right]=\frac{3}{6}$
$\text{P}\left(\text{E}\cup\text{F}|\text{G}\right)=\frac{\text{P}\left[\left(\text{E}\ \cup\ \text{F}\right)\cap\ \text{G}\right]}{\text{P}\left(\text{G}\right)}=\frac{\frac{3}{6}}{\frac{4}{6}}=\frac{3}{4}$
$\text{Again}\ \ \ \ \text{E}\cap\text{F}=\left(3\right)$
$\left(\text{E}\cap\text{F}\right)\cap\text{G}=(3)\ \Rightarrow\ \ \ \ \ \text{n}\big[\left(\text{E}\cap\text{F}\right)\cap\text{G}\big]=1$
$\text{P}\left[\left(\text{E}\ \cap\ \text{F}\right)\cap\text{G}\right]=\frac{1}{6}$
$\text{P}\left(\text{E}\cap\text{F}|\text{G}\right)=\frac{\text{P}\big[\left(\text{E}\ \cap\ \text{F}\right)\cap\ \text{G}\big]}{\text{P}\left(\text{G}\right)}=\frac{\frac{1}{6}}{\frac{4}{6}}=\frac{1}{4}$
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Question 1003 Marks
A lot of 100 watches is known to have 10 defective watches. If 8 watches are selected (one by one with replacement) at random, what is the probability that there will be at least one defective watch?
Answer
Probability of defective watch from a lot of 100 watches $=\frac{10}{100}=\frac{1}{10}$
$\therefore\text{p}=\frac{1}{10},\text{q}=\frac{9}{10},\text{n}=8$ and $\text{r}\geq1$
$\therefore\text{P}(\text{r}\geq1)=1-\text{P}(\text{r}=0)$
$=1-{^8}\text{C}_0\Big(\frac{1}{10}\Big)^0\Big(\frac{9}{10}\Big)^{8-0}$
$=1-\frac{8!}{0!8!}\Big(\frac{9}{10}\Big)^{8}=1-\Big(\frac{9}{10}\Big)^8$
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3 Marks Question - Page 2 - MATHS STD 12 Science Questions - Vidyadip