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MATHS 3 Marks Question

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

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Question 13 Marks
Find the mean $\mu$  variance $\sigma ^2$ for the following probability distribution:
X 0 1 2 3
P(X) $\frac{1}{6}$ $\frac{1}{2}$ $\frac{3}{10}$ $\frac{1}{30}$
Answer
$X:$ 0 1 2 3 Sum
$P(X):$ $\frac{1}{6}$ $\frac{1}{2}$ $\frac{3}{10}$ $\frac{1}{30}$ 1
$X.P(X):$ 0 $\frac{1}{2}$ $\frac{3}{5}$ $\frac{1}{10}$ $\frac{6}{5}$
$X^2P(X):$ 0 $\frac{1}{2}$ $\frac{6}{5}$ $\frac{1}{10}$ 2
Mean ($\mu$) = $\Sigma$X P(X)= $\frac{6}{5}$ or 1.2
Variance ($\sigma$)=$\Sigma X^2. P(X)-[\Sigma X. P(X)]^2 = \frac{14}{25}$ OR $0.56$.
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Question 23 Marks
Two dice are thrown together. What is the probability that the sum of the numbers on the two dice is neither 9 nor 11?
Answer
n (s) = 36P (getting the sum of 9) = P{(3,6), (6, 3), (4, 5), (5, 4)} = $\frac{4}{36}=\frac{1}{9}$
P (getting the sum of 11) = P {(5, 6), (6, 5)} = $\frac{2}{36}=\frac{1}{18}$
P (neither 9 or 11) = 1 – P (either 9 or 11) = 1 - $\Big[\frac{1}{9}+\frac{1}{18}\Big]$
= 1 - $\frac{3}{18}=1-\frac{1}{6}=\frac{5}{6}$.
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Question 33 Marks
There are two bags I and II. Bag I contains $3$ white and $4$ red balls and Bag II contains $5$ white and $6$ red balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag II.
Answer
Let $E_1$​​​​​​​: drawing a ball from bag I, $\therefore\text{P(E}_{1})=\frac{1}{2}$
$E_2$​​​​​​​: drawing a ball from bag II, $\therefore\text{P(E}_{2})=\frac{1}{2}$
A: getting a red ball, $\therefore\text{P(A/E}_{1})=\frac{4}{7},\text{ P(A/E}_{2})=\frac{6}{11}$
$\therefore$ Required probability = $\therefore\text{ P(E}_{2}/\text{A})=\frac{\text{P(E}_{2})\cdot\text{P(A/E}_{2})}{\text{P(E}_{1})\cdot\text{P(A/E}_{1})+\text{P(E}_{2})\cdot\text{P(A/E}_{2})}$
$\frac{\frac{1}{2}\cdot\frac{6}{11}}{\frac{1}{2}\cdot\frac{4}{7}+\frac{1}{2}\cdot\frac{6}{11}}=\frac{21}{43}$.
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Question 43 Marks
Determine the binomial distribution for which the mean is $20$ and variance $16$.
Answer
$n. p = 20, n p q = 16$
$\Rightarrow\text{q}=\frac{16}{20}=\frac{4}{5}\text{ }\text{ }\therefore\text{ }\text{ }\text{p}=\frac{1}{5}\text{and n}=100$
OR The Distribution is P(r) = $100_{\text{C}_\text{r}}\Bigg[\frac{4}{5}\Bigg]^\text{100 - r}\cdot\Bigg(\frac{1}{5}\Bigg)^\text{r},\text{ r = 0,1,2,.....,100.}$
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Question 53 Marks
An urn contains 4 red and 7 blue balls. Two balls are drawn at random with replacement. Find the probability of getting:
  1. 2 red balls.
  2. 2 blue balls.
  3. One red and one blue ball.
Answer
  1. P (2 red balls) =$\frac{4}{11}\cdot\frac{4}{11}=\frac{16}{121}$.
  2. P (2 blue balls) = $\frac{7}{11}\cdot\frac{7}{11}=\frac{49}{121}$.
  3. P (one red and one blue ball) = $\frac{4}{11}\cdot\frac{7}{11}\cdot2=\frac{56}{121}$.
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Question 63 Marks
A and B toss a coin alternately till one of them gets a head and wins the game. If A starts first, find the probability that B will win the game.
Answer
P (Head) = $\frac{1}{2}$ P (Tail) = $\frac{1}{2}$
P (B wins) = $\text{P}[(\overline{\text{A}}\text{B})\text{ OR }(\overline{\text{A}}{\text{ }}\overline{\text{B}}\text{ } \overline{\text{A}}{\text{ }}\text{B})\text{ OR }(\overline{\text{A}}{\text{ }}\overline{\text{B}}\text{ } \overline{\text{A}}{\text{ }}\overline{\text{B}}\text{ } \overline{\text{A}}{\text{ }}\text{B})\text{ OR }.......]$
$=\Bigg(\frac{1}{2}\cdot\frac{1}{2}\Bigg)+\Bigg(\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\Bigg)+\Bigg(\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\Bigg)+..........$
$=\frac{1}{4}\cdot\Bigg(1+\frac{1}{4}+\Bigg(\frac{1}{4}\Bigg)^{2}+.......\Bigg)=\frac{1}{4}\Bigg(\frac{1}{1-\frac{1}{4}}\Bigg)=\frac{1}{4}\cdot\frac{4}{3}=\frac{1}{3}.$
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Question 73 Marks
Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of number of jacks.
Answer
$\text{p}=\frac{4}{52}=\frac{1}{13},\text{ q}=\frac{12}{13}$
Number of Jacks (X): 0 1. 2.
P(X): $\Bigg(\frac{12}{13}\Bigg)^{2}$ $2.\frac{1}{13}\cdot\frac{12}{13}$ $\Bigg(\frac{1}{13}\Bigg)^{2}$
  $\frac{144}{169}$ $\frac{24}{169}$ $\frac{1}{169}$
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Question 83 Marks
There are two bags I and II. Bag I contains 2 white and 3 red balls and Bag II contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag II.
Answer
Let $\text{E}_{1} :$ drawing ball from Bag I $\Rightarrow P(E_{1}) = \frac{1}{2}$$\text{E}_{2} :$ drawing a ball from Bag II $\Rightarrow P E_{2} = \frac{1}{2}$
$\text{A} :$ Getting a red ball $\Rightarrow P \bigg(\frac{A}{E_1}\bigg) = \frac{3}{5}, P\bigg( \frac{A}{E_{2}} \bigg) = \frac{5}{9}$
$P \bigg( \frac{E_{2}}{A}\bigg) =\frac{P(E_{2}).\bigg(\frac{A}{E_2}\bigg)}{P (E_{1}). P\bigg(\frac{A}{E_{1}}\bigg)+ P \bigg( P(E_{2}) . P\bigg(\frac{A}{E_{2}}\bigg)}$
$= \frac{\frac{1}{2}.\frac{5}{9}}{{\frac{1}{2}}.\frac{3}{5} + \frac{1}{2} .\frac{5}{9}} = \frac{25}{52}$
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Question 93 Marks
A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability that it is neither an ace nor a king.
Answer
$\text{P (an Acc)} =\frac{4}{52} = \frac{1}{13}$$\text{P (a king)} =\frac{4}{52} = \frac{1}{13}$
$\therefore \text{P(a king or an Ace)} = \frac{1}{13} + \frac{1}{13} = \frac{2}{13}$
$\therefore \text{P(Neither a king nor an Ace)} = 1 -\frac{2}{13} = \frac{11}{13}$
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Question 103 Marks
Find the binomial distribution for which the mean is 4 and variance 3.
Answer
$\text{np} = \text{4 and npq} = 3 $$\therefore \text{q} = \frac{3}{4} \Rightarrow \text{p} = \frac{1}{4} $
$\text{np} = 4 \Rightarrow \text{n} \times\frac{1}{4}= 4 \Rightarrow \text{n} = 16$
$\text{For writing n} = 16, \text{p} = \frac{1}{4}, \text{q} = \frac{3}{4} \text{or P (r)} = 16_\stackrel{{C}}{{r}}\bigg(\frac{1}{4}\bigg)^{16 - r} , r = 0, 1, 2, \dots\dots16$
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Question 113 Marks
Find mean $u,$variance $\sigma^{2}$for the following probability distribution:
X 0 1 2 3
P(X) $\frac{1}{8}$ $\frac{3}{8}$ $\frac{3}{8}$ $\frac{1}{8}$
Answer
$\text{X}{_i}$ 0 1 2 3 Total
$\text{P (X}_{i})$ $\frac{1}{8}$ $\frac{3}{8}$ $\frac{3}{8}$ $\frac{1}{8}$ 1
$\text{X}_{i} \text{P X}_{i}$ 0 $\frac{3}{8}$ $\frac{6}{8}$ $\frac{3}{8}$ $\frac{3}{2}$
$\text{Xi}^{2} \text{P (X}_{i})$ 0 $\frac{3}{8}$ $\frac{3}{2}$ $\frac{3}{9}$ $\frac{24}{8} = 3$
$\text{Mean (u)} \Sigma \text{X}_{i} \text{P X}{i} = \frac{3}{2} \text{or 1.5} $
$\text{Variance} (\sigma^{2}) = \Sigma X^{2}_{i} P(X_{i}) - [\Sigma X _{i} P(X_{i}) ]^{2} $
$= 3 - \bigg(\frac{3}{2}\bigg)^{2}$
$3 - \frac{9}{4} = \frac{3}{4} \text{or 0.75}$
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Question 123 Marks
A pair of dice is tossed twice. If the random variable X is defined as the number of doublets, find the probability distribution of X.
Answer
Let x denote the number of doublets. Possible doublets are: -(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
Clearly X can take values 0, 1, 2
Probability of:-
  1. Getting a doublet $= \frac{1}{6}$
  2. Not getting a doublet $= 1 -\frac{1}{6} = \frac{5}{6}$
P (X = 0) = P (not doublet) $= \frac{5}{6}\times \frac{5}{6} = \frac{25}{36}$
P(X = 1) = P (one doublet and one not doubled) $= 2 \frac{1}{6} \frac{5}{6} = \frac{10}{36}$
P(X = 2) = P (both doublets) $= \frac{1}{6} \frac{1}{6} = \frac{1}{36}$
$\therefore$ Required probability distribution is:-
X
0
1
2
P (X)
$\frac{25}{36}$
$\frac{10}{36}$
$\frac{1}{36}$
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Question 133 Marks
In a factory, which manufactures nuts, machines A, B and C manufacture respectively 25%, 35% and 40% of nuts. Of their outputs, 5, 4 and 2 per cent respectively are defective nuts. A nut is drawn at random from the product and is found to be defective. Find the probability that it is manufactured by machine B.
Answer
$\text{Let the events}A_{1}, A_{2}, A_{3} \text{E be as follows} :$$A_{1} :$ the nut is manufactured by machine A
$A_{2} :$ the nut is manufactured by machine B
$A_{3} :$ the nut is manufactured by machine C
${E:}$ the nut manufactured is defective
$\therefore P(A_{1}) = 0.25, P(A_{2}) = 0.35, P(A_{3}) = 0.40 $
$P (E/A_{1}) = 0.05, P(E/A_{2}) = 0.04, P(E/A_{3}) = 0.02$
$P(A_{2}/E) = \frac{P(A_{2}).P(E/A_{2})}{\sum\limits^{3}_{i-1}P(A_{i}).P(E/A_i)}$
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Question 143 Marks
If the mean and variance of a binomial distribution are respectively 9 and 6, find the distribution.
Answer
$\text{Here Mean = np = 9 and variance = npq = 6}$$\therefore q = \frac{6}{9} = \frac{2}{3}$
$\therefore p = 1- q= \frac{1}{3}$
$\text{Again np} = 9$
$\Rightarrow n \times \frac{1}{3} = 9 \Rightarrow n = 27$
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Question 153 Marks
Two dice are rolled once. Find the probability that:
  1. the numbers on two dice are different.
  2. the total of numbers on the two dice is at least.
Answer
P(Numbers on two dice are different) $= \frac{30}{36} =\frac {5}{6}$ P (Total of numbers on two dice is atleast 4) $=\frac{33}{36} = \frac{11}{12}$
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Question 163 Marks
8% of people in a group are left handed. What is the probability that 2 or more of a random sample of 25 from the group are left handed? $e^{-2} = 0.135$
Answer
Here ${\lambda} = np = 25 \times \frac{8}{100} =2 $$\text{Required probability} = P (X\geq{-} 2)$
$= 1 -P (X = 0) - P (X = 1)$
$= 1-\frac{e^{-2}}{1!} - \frac{2e^{-2}}{1!}$
$= 1 - 3e^{-2} = 1 - 3(.135) = 0.595$
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Question 173 Marks
An instructor has a test bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the test bank, what is the probability that it will be an easy question given that it is a multiple choice question?
Answer
Number of easy True/False questions = 300
Number of difficult True/False questions = 200
Number of easy multiple choice questions = 500
Number of difficult multiple choice questions = 400
Total number of all such questions = n(S) = 1400
Let E represents an easy question and F represents a multiple choice question.
$\therefore\ \text{n}(\text{E})\ =300+500=800$
$\text{and}\ \text{n}(\text{F})\ =500+400=900$
$\text{P}(\text{F})=\frac{\text{n}(\text{F})}{\text{n}(\text{S})}=\frac{900}{1400}$
$\text{n}(\text{E}\cap\text{F})=500\ \Rightarrow\ \ \ \ \ \text{P}(\text{E}\cap\text{F})=\frac{\text{n}(\text{E}\cap\text{F})}{\text{n}(\text{S})}=\frac{500}{1400}$
$\text{P}(\text{E}|\text{F})=\frac{\text{P}(\text{E}\cap\text{F})}{\text{P}(\text{F})}=\frac{\frac{500}{1400}}{\frac{900}{1400}}=\frac{500}{900}=\frac{5}{9}$
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Question 183 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 193 Marks
On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?
Answer
The repeated guessing of correct answers from 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 n = 5 and $\text{p}=\frac{1}{3}$
$\therefore\ \text{P}(\text{X=x})=\ ^\text{n}\text{C}_\text{x}\text{q}^\text{n-x}\text{p}^\text{x}$
$=\ ^5\text{C}_\text{x}\bigg(\frac{2}{3}\bigg)^{5-\text{x}}.\bigg(\frac{1}{3}\bigg)^\text{x}$
P(guessing more than 4 correct answers) = P(X ≥ 4)
$=\text{P}(\text{X}=4)+\text{P}(\text{X}=5)$
$=\ ^5\text{C}_\text{4}\bigg(\frac{2}{3}\bigg).\bigg(\frac{1}{3}\bigg)^4+\ ^5\text{C}_\text{5}\bigg(\frac{1}{3}\bigg)^5$
$=5\cdot\frac{2}{3}\cdot\frac{1}{81}+1\cdot\frac{1}{243}$
$=\frac{10}{243}+\frac{1}{243}$
$=\frac{11}{243}$
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Question 203 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 213 Marks
A dice is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?
Answer
The sample space of the experiment is {(1, 1), (1, 2), (1, 3), ....., (6, 6)} conisting of 36 outcomes.
$\text{P(A)}=\text{P}(\text{Sum}=6)=\frac{5}{36}$
$\text{P(B)}=\text{P}(4\text{ appears at least once})=\frac{11}{36}$
Now, $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A and B})}{\text{P(A)}}$
$=\frac{\text{P}(\text{Sum is } 6 \text{ and }4 \text{ has appeard at least once})}{\text{P(A)}}$
$=\frac{\frac{2}{36}}{\frac{5}{36}}$
$=\frac{2}{5}$
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Question 223 Marks
If the sum of the mean and variance of a binomial distribution for 6 trials is $\frac{10}{9},$ find the distribution.
Answer
Given that $\text{n}=6$
The sum of mean and variance of a binomial distribution for 6 trials is $\frac{10}{3}.$
$\Rightarrow6\text{p}+6\text{pq}=\frac{10}{3}$
$\Rightarrow18\text{p}+18\text{p}(1-\text{p})=10$
$\Rightarrow18\text{p}^2-36\text{p}+10=0$
$\Rightarrow(3\text{p}-1)(6\text{p}-10)=0$
$\Rightarrow\text{p}=\frac{1}{3}$ or $\frac{5}{3}$
$\text{p}=\frac{5}{3}$ (neglected as it is greater than 1)
$\therefore\text{p}=\frac{1}{3}$
$\Rightarrow\text{q}=1-\text{p}=\frac{2}{3}$
Hence, the distribution is given by
$\text{P(X = r})\text{ }^6\text{C}_{\text{r}}\big(\frac{1}{3}\big)^{\text{r}}\big(\frac{2}{3}\big)^{6-\text{r}},\text{r}=0,1,2\dots6$
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Question 233 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 243 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 $=\mathrm{np}$ varience $=\mathrm{npq}$ Mean - variance $=\mathrm{np}-\mathrm{npq}=\mathrm{np}(1$ $-q)=n p . p=n p^2$ Mean - variance $>0$ Mean $>$ variance
So, mean can never be less than variance.
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Question 253 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 263 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 273 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 283 Marks
Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostlier?
Answer
Let $\text{E}_1,\ \text{E}_2$ E be the events$\text{E}_1$ : 'A student is residing in hostel'
$\text{E}_2$: 'A student is day scholar',
E : 'A student gets A grade',
$\text{Now},\ \ \ \ \ \text{P}(\text{E}_1)=\frac{60}{100}=\frac{3}{5}\ \text{and}\ \text{P}(\text{E}_2)=\frac{40}{100}=\frac{2}{5}$
$\therefore\ \text{P}(\text{E}|\text{E}_1)=\frac{30}{100}=\frac{3}{10}\ \text{and}\ \text{P}(\text{E}|\text{E}_2)=\frac{20}{100}=\frac{2}{10}$
Required probability $=\text{P}(\text{E}_1|\text{E})$
$=\frac{\text{P}(\text{E}_1)\text{P}(\text{E}|\text{E}_1)}{\text{P}(\text{E}_1)\text{P}(\text{E}|\text{E}_1)+\text{P}(\text{E}_2)\text{P}(\text{E}|\text{E}_2)}$ (By Baye's Theorem)
$=\frac{\frac{3}{5}\times\frac{3}{10}}{\frac{3}{5}\times\frac{3}{10}+\frac{2}{5}\times\frac{2}{10}}=\frac{9}{9+4}=\frac{9}{13}.$
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Question 293 Marks
If the probability of a defective bolt is 0.1, find the (1) mean and (2) standard deviation for the distribution of bolts in a total of 400 bolts.
Answer
Let p denotes the probability of selecting a defective bolt, so
$\text{p}=0.1$
$\text{p}=\frac{1}{10}$
$\text{q}=1-\frac{1}{10}$ [Since p + q = 1]
$\text{q}=-\frac{9}{10}$
Given, $\text{n}=400$
(1)
$\text{Mean = np}$
$=400\times\frac{1}{10}$
$\text{Mean = 40}$
(2)
$\text{Standard deviation} = \sqrt{\text{npq}}$
$=\sqrt{400\times\frac{1}{10}\times\frac{9}{10}}$
$=\sqrt{36}$
$\text{standard deviation}=6$
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Question 303 Marks
A coin is tossed 5 times. What is the probability that tail appears an odd number of times?
Answer
Let X denote the number of tails when a coin is tossed 5 times.
X follows a binomial distribution with $\text{n}=5;\text{p}=\frac{1}{2};\text{q}=1-\text{p}=\frac{1}{2}$
Then $\text{P}(\text{X = r})=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{\text{n}-\text{r}}=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{2}\big)^5$
The required probability $=\text{P}(\text{X = odd})$
$=\text{P}(\text{X}=1)+\text{P}(\text{X}=3)+\text{P}(\text{X}=5)$
$=\text{ }^5\text{C}_1\big(\frac{1}{2}\big)^5+\text{ }^5\text{C}_3\big(\frac{1}{2}\big)^5+\text{ }^5\text{C}_5\big(\frac{1}{2}\big)^5$
$=\big(\frac{1}{2}\big)^5\ [5+10+1]$
$=\frac{16}{32}$
$=\frac{1}{2}$
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Question 313 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 323 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 333 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 343 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 353 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 363 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 373 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 383 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 393 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 403 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 413 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 423 Marks
A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number is even,’ and B be the event, ‘the number is red’. Are A and B independent?
Answer
When a die is thrown, the sample space (S) is$\text{S}=\{1,\ 2,\ 3,\ 4,\ 5,\ 6\}$
Let A: the number is even = {2, 4, 6}
$\Rightarrow\text{P}(\text{A})=\frac{3}{6}=\frac{1}{2}$
B: the number is red = {1, 2, 3}
$\Rightarrow\text{P}(\text{B})=\frac{3}{6}=\frac{1}{3}$
$\therefore\text{A}\cap\text{B}=\left\{2\right\}$
$\text{P}(\text{AB})=\text{P}(\text{A}\cap\text{B})=\frac{1}{6}$
$\text{P}(\text{A}).\text{P}(\text{B})=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}\neq\frac{1}{6}$
$\Rightarrow\text{P}(\text{A}).\text{P}(\text{B})\neq\text{P}(\text{AB})$
Therefore, A and B are not independent.
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Question 433 Marks
Determine P(E|F) : A coin is tossed three times.
E : heads on third toss, F : heads on first two tosses.
Answer
A coin tossed three times, i.e.,$\text{S}=(\text{TTT, HTT, THT, TTH, HHT, HTH, THH, HHH})$$\ \ \ \ \ \Rightarrow\ \ \ \ \ \ \ \text{n}(\text{S})=8$
E : heads on third toss
$\text{E}=(\text{TTH, HTH, THH, HHH})$$\ \ \ \ \ \ \ \ \Rightarrow\ \ \ \ \ \text{n}(\text{E})=4$
$\text{P}\left(\text{E}\right)=\frac{\text{n}\left(\text{E}\right)}{\text{n}\left(\text{S}\right)}=\frac{4}{8}=\frac{1}{2}$
F : heads on first two tosses
$\text{F}(\text{HHT, HHH})$$\ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow\ \ \ \ \ \ \ \ \text{n}(\text{F})=2$
$ \text{P}\left(\text{F}\right)=\frac{\text{n}\left(\text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{2}{8}=\frac{1}{4}$
$\therefore\ \ \ \ \ \ \ \ \text{E}\cap\text{F}=\left(\text{HHH}\right)\ \ \ \ \ \ \Rightarrow\text{n}\left(\text{E}\cap\text{F}\right)=1$
$\therefore\ \ \ \ \ \ \ \ \text{P}\left(\text{E}\cap\text{F}\right)=\frac{\text{n}\left(\text{E}\cap\text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{1}{8}$
$\text{And}\ \ \ \ \text{P}\left(\text{E}|\text{F}\right)=\frac{\text{P}\left(\text{E}\cap\text{F}\right)}{\text{P}\left(\text{F}\right)}=\frac{\frac{1}{8}}{\frac{2}{8}}=\frac{1}{2}$
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Question 443 Marks
A pair of dice is thrown 6 times. If getting a total of 9 is considered a success, what is the probability of at least 5 successes?
Answer
Let X be the number of successes in 6 throws of the two dice. Probability of success = Probability of getting a total of 9 = Probability of getting (3, 6), (4, 5), (5, 4), (6, 3) out of 36 outcomes $\text{p}=\frac{4}{36}=\frac{1}{9},\text{q}=1-\text{p}=\frac{8}{9}\text{ and }\text{n}=6$ X follows a binomial distribution with $\text{n}=6,\text{p}=\frac{1}{9}$ and $\text{q}=\frac{8}{9}$$\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 453 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 463 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 473 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 483 Marks
Find the mean and standard deviation of the following probability distributions:
$\text{x}_\text{i}$
$0$
$1$
$2$
$3$
$4$
$5$
$\text{p}_\text{i}$
$\frac{1}{6}$
$\frac{5}{18}$
$\frac{2}{9}$
$\frac{1}{6}$
$\frac{1}{9}$
$\frac{1}{18}$
Answer
$\text{x}_\text{i}$ $\text{p}_\text{i}$ $\text{p}_\text{i}\text{x}_\text{i}$ $\text{p}_\text{i}\text{x}_\text{i}^2$
$0$ $\frac{1}{6}$ $0$ $0$
$1$ $\frac{5}{18}$ $\frac{5}{18}$ $\frac{5}{18}$
$2$ $\frac{2}{9}$ $\frac{4}{9}$ $\frac{8}{9}$
$3$ $\frac{1}{6}$ $\frac{1}{2}$ $\frac{3}{2}$
$4$ $\frac{1}{9}$ $\frac{4}{9}$ $\frac{16}{9}$
$5$ $\frac{1}{18}$ $\frac{5}{18}$ $\frac{25}{18}$
    $\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{35}{18}$ $\sum\text{p}_\text{i}\text{x}_\text{i}^2=\frac{35}{6}$
Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{35}{18}$
Variance $\sum\text{p}_\text{i}\text{x}_\text{i}^2-\big(\sum\text{p}_\text{i}\text{x}_\text{i}\big)^2=\frac{35}{6}-\Big(\frac{35}{18}\Big)^2=\frac{665}{324}$
Standard Deviation $=\sqrt{\text{Variation}}=\frac{\sqrt{665}}{18}$
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Question 493 Marks
Find the probability of getting 5 exactly twice in 7 throws of a die.
Answer
The repeated tossing of a die are Bernoulli trials. Let X represent the number of times of getting 5 in 7 throws of the die.
Probability of getting 5 in a single throw of the die, $\text{p}=\frac{1}{6}$
$\therefore\ \text{q}=1-\text{p}=1-\frac{1}{6}=\frac{5}{6}$
Clearly, X has the probability distribution with n = 7 and $\text{p}=\frac{1}{6}$
$\therefore\ \text{P}(\text{X=x})=\ ^\text{n}\text{C}_\text{x}\text{q}^\text{n-x}\text{p}^\text{x}=\ ^7\text{C}_\text{x}\bigg (\frac{5}{6}\bigg)^{7-\text{x}}.\bigg (\frac{1}{6}\bigg)^\text{x}$
P(getting 5 exactly twice) = P(X = 2)
$=\ ^7\text{C}_2\bigg (\frac{5}{6}\bigg)^5.\bigg (\frac{1}{6}\bigg)^{2}$
$=21\cdot\Big(\frac{5}{6}\Big)^5\cdot\frac{1}{36}$
$=\Big(\frac{7}{12}\Big)\Big(\frac{5}{6}\Big)^5$
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Question 503 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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