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MATHS 5 Marks Questions

Mean and variance of a random variable · MP Board (MPBSE) STD 12 Science (Madhya Pradesh - English Medium). 27 questions.

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Question 15 Marks
Two cards are selected at random from a box which contains five cards numbered 1, 1, 2, 2, and 3. Let X denote the sum and Y the maximum of the two numbers drawn. Find the probability distribution, mean and variance of X and Y.
Answer
Box contains five cards $1, 1, 2, 2, 3$. Here,
X denotes the sum of the two number on cards drawn. Y denotes the maximum of the two number.
So, $X = 2, 3, 4, 5\ Y = 1, 2, 3\ P(X = 2) = P(1)P(1)$
$=\frac{2}{5}\times\frac{1}{4}$
$=0.1$ $ P(X = 3) = P(1)P(2) + P(2)P(1)$
$=\frac{2}{5}\times\frac{2}{4}+\frac{2}{5}\times\frac{2}{4}$
$=0.4$ $P(X = 4) = P(2)P(2) + P(1)P(3) + P(3)P(1)$
$=\frac{2}{5}\times\frac{1}{4}+\frac{2}{5}\times\frac{1}{4}+\frac{1}{5}\times\frac{2}{4}$
$=0.3$ $P(X = 5) = P(2)P(3) + P(3)P(2)$
$=\frac{2}{5}\times\frac{2}{4}+\frac{1}{5}\times\frac{2}{4}$
$=0.2$ Probability distribution for X
$x:$ $2$ $3$ $4$ $5$
$P(x):$ $0.1$ $0.4$ $0.3$ $0.2$
Now,
$x_i$ $p_i$ $p_i$ $x_ip_i$ ${x_i}^2p_i$
$2$ $0.1$ $0.1$ $0.4$
$3$ $0.4$ $1.2$ $3.6$
$4$ $0.3$ $1.2$ $4.8$
$5$ $0.2$ $1.0$ $5.0$
    $\sum\text{xp}=3.6$ $\sum\text{x}^2\text{p}=13.8$
Mean $=\sum\text{xp}$
Mean $= 3.6$
Variance $=\sum\text{x}^2\text{p}-(\text{mean})^2$
$=13.8-(3.6)^2$ $=13.8-12.96$
Variance $= 0.84\ P(Y = 1) = P(1)P(1)$
$=\frac{2}{5}\times\frac{1}{4}$
$=\frac{2}{20}$
$=0.1$ $P(Y = 2) = P(1)P(2) + P(2)P(1) + P(2)P(2)$
$=\frac{2}{5}\times\frac{2}{4}+\frac{2}{5}\times\frac{2}{4}+\frac{2}{5}\times\frac{1}{4}$
$=0.5$ $P(Y = 3) = P(1)P(3) + P(2)P(3) + P(3)P(1) + P(3)P(2)$
$=\frac{2}{5}\times\frac{1}{4}+\frac{2}{5}\times\frac{1}{4}+\frac{1}{5}\times\frac{2}{4}+\frac{1}{5}\times\frac{2}{4}$
$=0.4$ Probability distribution for Y is
$x:$
$1$
$2$
$3$
$p(x):$
$0.1$
$0.5$
$0.4$
 
$y_i$ $p_i$ $y_ip_i$ ${y_i}^2p_i$
$1$ $0.1$ $0.1$ $0.1$
$2$ $0.5$ $1.0$ $2.0$
$3$ $0.4$ $1.2$ $3.6$
    $\sum\text{xp}=2.3$ $\sum\text{x}^2\text{p}=5.7$
Mean $=\sum\text{xp}=2.3$ Variance $=\sum\text{x}^2\text{p}-(\text{mean})^2$ $=5.1-(2.3)^2$ Variance $= 0.41$
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Question 25 Marks
Two numbers are selected at random (without replacement) from positive integers 2, 3, 4, 5, 6 and 7. Let X denote the larger of the two numbers obtained. Find the mean and variance of the probability distribution of X.
Answer
We can select two positive in 6 × 5 = 30 different waysP(X = 3) = P(larger number is 3) = {(2, 3), (3, 2)} $=\frac{2}{30}$
P(X = 4) = P(larger number is 4) = {(2, 4), (4, 2), (3, 4), (4, 3)} $=\frac{4}{30}$
P(X = 5) = P(larger number is 5) = {(2, 5), (5, 2), (3, 5), (5, 3), (4, 5), (5, 4)} $=\frac{6}{30}$
P(X = 6) = P(larger number is 6) = {(2, 6), (6, 2), (3, 6), (6, 3), (4, 6), (6, 4), (5, 6), (6, 5)} $=\frac{8}{30}$
P(X = 7) = P(larger number is 7) = {(2, 7), (7, 2), (3, 7), (7, 3), (4, 7), (7, 4), (5, 7), (7, 5), (6, 7), (7, 6)} $=\frac{10}{30}$
Thus, the probability distribution of random variable X is,
$\text{x}_\text{i}$ $\text{p}_\text{i}$ $\text{x}_\text{i}\text{p}_\text{i}$ $\text{x}_\text{i}^2\text{p}_\text{i}$
$3$ $\frac{2}{30}$ $\frac{6}{30}$ $\frac{18}{30}$
$4$ $\frac{4}{30}$ $\frac{16}{30}$ $\frac{64}{30}$
$5$ $\frac{6}{30}$ $\frac{30}{30}$ $\frac{150}{30}$
$6$ $\frac{8}{30}$ $\frac{48}{30}$ $\frac{288}{30}$
$7$ $\frac{10}{30}$ $\frac{70}{30}$ $\frac{490}{30}$
    $\sum\text{x}_\text{i}\text{p}_\text{i}=\frac{17}{3}$ $\sum\text{x}_\text{i}\text{p}_\text{i}^2=\frac{101}{3}$
Mean $=\sum\text{x}_\text{i}\text{p}_\text{i}=\frac{17}{3}$
Variance $=\sum\text{x}_\text{i}\text{p}_\text{i}-\big(\sum\text{x}_\text{i}\text{p}_\text{i}\big)^2=\frac{101}{3}-\Big(\frac{17}{3}\Big)=\frac{14}{9}$
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Question 35 Marks
In a game, a man wins Rs 5 for getting a number greater than 4 and loses Rs 1 otherwise, when a fair die is thrown. The man decided to thrown a die thrice but to quit as and when he gets a number greater then 4. Find the expected value of the amount he wins or loses.
Answer
The man may get number greater than 4 in the first throw and then he quits the game. He may get a number less than equat to 4. In the first throw and in the second throw he may get the number greater than 4 and quits the game.
In the first two throws he gets a number less than equal to 4 and in the third throw he may get a number greater than 6. He may not get number greater than 4 in any one of three throws.
Let X be the amount he wins/looses.
Then, X can take values -3, 3, 4, 5 such that
P(X= 5) = P(Getting number greater than 4 in first throw) $=\frac{1}{3}$
(X= 4) = P(Getting number less than equal to 4 in the first throw and number greater than 4 in second throw) $=\frac{4}{6}\times\frac{2}{6}=\frac{2}{9}$
P (X = 3) = P(Getting number less than equal to 4 in the first two throws and number greater than 4 in third throw) $=\frac{4}{6}\times\frac{4}{6}\times\frac{2}{6}=\frac{4}{27}$
P(X = -3) = P(Getting number less than equal to 4 in all three throws) $=\frac{4}{6}\times\frac{4}{6}\times\frac{4}{6}=\frac{8}{27}$
$\text{X}$ $5$ $4$ $3$ $-3$
$\text{P}(\text{X})$ $\frac{1}{3}$ $\frac{2}{9}$ $\frac{4}{27}$ $\frac{8}{27}$
$\text{E}(\text{X})=\Big(5\times\frac{1}{3}\Big)+4\Big(\frac{2}{9}\Big)+3\Big(\frac{4}{27}\Big)-3\Big(\frac{8}{27}\Big)$
$=\frac{1}{27}(45+24+12-24)$
$=\frac{57}{27}$
$=\frac{19}{9}$
Expected value of the amount he wins or loses is $=\frac{19}{9}$
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Question 45 Marks
Let X denot the number of colleges where you will apply after your results and P(X = x) denotes your probability of getting admission in x number of colleges. It is given that
$\text{P}(\text{X = x})=\begin{cases}\text{kx},&\text{if}\text{ x}=0\text{ or }1\\2\text{kx},&\text{if x = 2}\\\text{k}(5-\text{x}),&\text{if x = 3 or 4}\\0,&\text{if x > 4}\end{cases}$
where k is a positive constant. Find the value of k. Also find the probability that you will get addmission in
  1. Exactly one college.
  2. At most two colleges.
  3. At least two colleges.
Answer
The probability distribution of X is
X 0 1 2 3 4
P(X) 0 k 4k 2k k
The given distribution is a probability distribution.
$\therefore\ \sum\text{p}_\text{i}=1$
⇒ 0 + k + 4k + 2k + k
⇒ 8k = 1
⇒ k = 0.125
  1. P(getting admission in exactly one college) $= \text{P}(\text{X }= 1) = \text{k} = 0.125$
  2. P(getting admission in at most 2 colleges) $=\text{P}(\text{X}\leq2)=0+\text{k}+4\text{k}=5\text{k}=0.625$
  3. P(getting admission in atleast 2 colleges) $=\text{P}(\text{X}\geq2)=4\text{k}+2\text{k}+\text{k}=7\text{k}=0.875$
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Question 55 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 65 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 75 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 85 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
$x_i$ $p_i$ $p_i$ $p_ix_i$ ${p_ix_i}^2$
$a$ $p$ $ap$ $a^2p$
$b$ $q$ $bq$ $b^2q$
    $\sum\text{p}_\text{i}\text{x}_\text{i}=\text{ap}+\text{bq}$ $\sum\text{p}_\text{i}\text{x}_\text{i}^2=\text{a}^2\text{p}+\text{b}^2\text{q}$
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 95 Marks
Two cards are drawn simultaneosly from a well shuffled deck of 52 cards. Find the probability distribution of the number of the successes, when getting a spade is considered a success.
Answer
Let X denote the numbers of spades in a sample of two cards drawn from a well shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
P(X = 0)
= P(no spade)
$=\frac{\text{}^{39}\text{C}_2}{\text{}^{52}\text{C}_2}$
$=\frac{741}{1326}$
$=\frac{19}{34}$
P(X = 1)
= P(1 spade)
$=\frac{\text{}^{13}\text{C}_1\times\text{}^{39}\text{C}_1}{\text{}^{52}\text{C}_2}$
$=\frac{507}{1326}$
$=\frac{13}{34}$
P(X = 2)
= P(2 spade)
$=\frac{\text{}^{13}\text{C}_2}{\text{}^{52}\text{C}_2}$
$=\frac{78}{1326}$
$=\frac{1}{17}$
Thus, the probability distribution of X is given by
$\text{X}$
$0$
$1$
$2$
$\text{P}(\text{X})$
$\frac{19}{34}$
$\frac{13}{34}$
$\frac{1}{17}$
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Question 105 Marks
A die is tossed twice. A 'success' is getting an odd number on a toss. Find the variance of the number of successes.
Answer
It is given that "success" denotes the event of getting the number 1, 3 or 5. Then, P(success) $=\frac{1}{2}$ Also, "failure" denotes the event of getting the numbers 2, 4, or 6. Then, P(failure) $=\frac{1}{2}$ Let X denote the event of getting success. Then, X can take the values 0, 1 and 2. Now, P(X = 0) = P(no success) $=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$ P(X = 1) = P(1 success) $=\Big(\frac{1}{2}\times\frac{1}{2}\Big)+\Big(\frac{1}{2}\times\frac{1}{2}\Big)=\frac{1}{2}$ P(X = 2) = P(2 success) $=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$ Thus, the probability distribution of X is given by
$\text{X}$ $\text{P}(\text{X})$
$0$ $\frac{1}{4}$
$1$ $\frac{1}{2}$
$2$ $\frac{1}{4}$
Computation of mean and variance
$\text{X}_\text{i}$ $\text{P}_\text{i}$ $\text{P}_\text{i}\text{X}_\text{i}$ $\text{P}_\text{i}\text{X}_\text{i}^2$
$0$ $\frac{1}{4}$ $0$ $0$
$1$ $\frac{1}{2}$ $\frac{1}{2}$ $\frac{1}{2}$
$2$ $\frac{1}{4}$ $\frac{1}{2}$ $1$
    $\sum\text{p}_\text{i}\text{x}_\text{i}=1$ $\sum\text{p}_\text{i}\text{x}_\text{i}^2=\frac{3}{2}$
Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=1$ Variance $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-({\text{Mean}})^2=\frac{3}{2}-1=\frac{1}{2}$
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Question 115 Marks
A fair coin is tossed four times. Let X denote the longest string of heads accuring. Find the probability distribution mean and variance of X.
Answer
Let the event of getting a head = H and getting a tail = T
Let X denote the variable longest consecutive heads accuring in 4 tosses. The possible values are
X = 0 (no heads) {T, T, T, T}
X = 1 (1 head) {H, T, T, T}
X = 2 (2 heads) {H, H, T, T}
X = 3 (3 heads) {H, H, H, T}
X = 4 (4 heads) {H, H, H, H}
n(S) = {(HHHH), (HHHT), (HHTT), (HTHH), (HTHT), (HTTH), (HTTT), (THHH), (THTH), (THHT), (THTT), (THTT), (TTHH), (TTHT), (TTTH), (TTTT)}
$\text{P}(\text{X}=0)=\frac{1}{16}$
$\text{P}(\text{X}=1)=\frac{7}{16}$
$\text{P}(\text{X}=2)=\frac{5}{16}$
$\text{P}(\text{X}=3)=\frac{2}{16}$
$\text{P}(\text{X}=4)=\frac{1}{16}$
Thus, the probability distribution is
$\text{X}$ $0$ $1$ $2$ $3$ $4$
$\text{p}_\text{i}=\text{P}(\text{X})$ $\frac{1}{16}$ $\frac{7}{16}$ $\frac{5}{16}$ $\frac{2}{16}$ $\frac{1}{16}$
$\text{p}_\text{i}\text{x}_\text{i}^2$ $0$ $\frac{7}{16}$ $\frac{20}{16}$ $\frac{18}{16}$ $1$
Mean $=\sum_\text{i=1 to n}\text{X}_\text{i}\times\text{P}(\text{X}_\text{i})$
Mean, $\mu=0\times\frac{1}{16}+1\times\frac{7}{16}+2\times\frac{5}{16}+3\times\frac{2}{16}+4\times\frac{1}{16}$
$=0+\frac{7}{16}+\frac{10}{16}+\frac{6}{16}+\frac{4}{16}$
$=\frac{27}{16}=1.7$
Variance Var(X) $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-\big(\sum\text{p}_\text{i}\text{x}_\text{i}\big)^2$
$=\frac{61}{16}-(1.7)^2$
$=3.825-2.89$
$=0.935$
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Question 125 Marks
Five defective mangoes are acciedently mixed with 15 good ones. Four mangoes are drawn at random from this lot. Find the probability distribution of the number of defective mangoes.
Answer
Let X denote number of defective mangoes in a sample of 4 mangoes drawn from a bag containing 5 defective mangoes and 15 good mangoes. Then, X can take the values 0, 1, 2, 3 and 4. Now, P(X = 0) P(no defective mango) $=\frac{\text{}^{15}\text{C}_4}{\text{}^{20}\text{C}_4}$ $=\frac{1365}{4845}$ $=\frac{91}{323}$ P(X = 1) = P(1 defective mango) $=\frac{\text{}^5\text{C}_1\times\text{}^{15}\text{C}_3}{\text{}^{20}\text{C}_4}$ $=\frac{2275}{4845}$ $=\frac{455}{969}$ P(X = 2) = P(2 defective mangoes) $=\frac{\text{}^5\text{C}_2\times\text{}^{15}\text{C}_2}{\text{}^{20}\text{C}_4}$ $=\frac{1050}{4845}$ $=\frac{70}{323}$ P(X = 2)= P(3 defective mangoes)
$=\frac{\text{}^5\text{C}_3\times\text{}^{15}\text{C}_1}{\text{}^{20}\text{C}_4}$ $=\frac{150}{4845}$ $=\frac{10}{323}$ P(X = 3)= P(4 defective mangoes)
$=\frac{\text{}^5\text{C}_4}{\text{}^{20}\text{C}_4}$ $=\frac{5}{4845}$ $=\frac{1}{969}$ The required probability distribution of X is given by
$\text{X}$
$0$
$1$
$2$
$3$
$4$
$\text{P}(\text{X})$
$\frac{91}{323}$
$\frac{455}{969}$
$\frac{70}{323}$
$\frac{10}{323}$
$\frac{1}{969}$
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Question 135 Marks
An urn contains 4 red and 3 blue balls. Find the probability distribution of the number of blue balls in a random draw of 3 balls with replacement.
Answer
Urn has 4 red and 3 blue balls. 3 balls are drawn with replacement. Let X denote the numbers of blue balls drawn out of 3 drawn.
So, X has values 0, 1, 2, 3
$\text{P}(\text{X}=0)=\text{P}\big(\overline{\text{B}}_1\big)\times\text{P}\big(\overline{\text{B}}_2\big)\times\text{P}\big(\overline{\text{B}}_3\big)$
$=\frac{4}{7}\times\frac{4}{7}\times\frac{4}{7}$
$=\frac{144}{343}$
$\text{P}(\text{X}=1)=\text{P}(\text{B}_1)\text{P}\big(\overline{\text{B}}_2\big)\text{P}\big(\overline{\text{B}}_3\big)+\text{P}\big(\overline{\text{B}}_1\big)\text{P}(\text{B}_2)\text{P}\big(\overline{\text{B}}_3\big)+\text{P}\big(\overline{\text{B}}_1\big)\text{P}\big(\overline{\text{B}}_2\big)\text{P}(\text{B}_3)$
$=\frac{3}{7}\times\frac{4}{7}\times\frac{4}{7}+\frac{4}{7}\times\frac{3}{7}\times\frac{4}{7}+\frac{4}{7}\times\frac{4}{7}\times\frac{3}{7}$
$=\frac{144}{343}$
$\text{P}(\text{X}=2)=\text{P}(\text{B}_1)\text{P}(\text{B}_2)\text{P}\big(\overline{\text{B}}_3\big)+\text{P}(\text{B}_1)\text{P}\big(\overline{\text{B}}_2\big)\text{P}(\text{B}_3)+\text{P}\big(\overline{\text{B}}_1\big)\text{P}(\text{B}_2)\text{P}(\text{B}_3)$
$=\frac{3}{7}\times\frac{3}{7}\times\frac{4}{7}+\frac{3}{7}\times\frac{4}{7}\times\frac{3}{7}+\frac{4}{7}\times\frac{3}{7}\times\frac{3}{7}$
$=\frac{108}{343}$
$\text{P}(\text{X}=3)=\text{P}(\text{B}_1)\text{P}(\text{B}_2)\text{P}(\text{B}_3)$
$=\frac{3}{7}\times\frac{3}{7}\times\frac{3}{7}$
$=\frac{27}{343}$
So,
Required probability distribution is
$\text{X}:$
$0$
$1$
$2$
$3$
$\text{P}(\text{X}):$
$\frac{64}{343}$
$\frac{144}{343}$
$\frac{108}{343}$
$\frac{27}{343}$
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Question 145 Marks
An urn contains 5 red and 2 blcak balls. Two balls are randomly drawn, without replacement. Let X represent the number of black balls drawn. What are the possible values of X? Is X a random variable? If yes, then find the mean and variance of X.
Answer
X can assume values 0 to 2. Yes X is a random variable. P(X = 0) = (Probability of getting no black ball) $=\frac{\text{}^{2}\text{C}_0\times\text{}^{5}\text{C}_2}{\text{}^{7}\text{C}_2}=\frac{1\times\frac{5\times4}{2\times1}}{\frac{7\times6}{2\times1}}=\frac{20}{42}$ P(X = 1) = (Probability of getting one black ball) $=\frac{\text{}^{2}\text{C}_1\times\text{}^{5}\text{C}_1}{\text{}^{7}\text{C}_2}=\frac{1\times5}{\frac{7\times6}{2\times1}}=\frac{20}{42}$ P(X = 2) = (Probability of getting two black balls) $=\frac{\text{}^{2}\text{C}_2\times\text{}^{5}\text{C}_0}{\text{}^{7}\text{C}_2}=\frac{1\times1}{\frac{7\times6}{2\times1}}=\frac{2}{42}$ Thus, Probability distribution of random variable X is
$\text{X}$
$0$
$1$
$2$
$\text{P}(\text{X}) $
$\frac{20}{42}$
$\frac{20}{42}$
$\frac{2}{42}$
$\text{x}_\text{i}$ $\text{p}_\text{i}$ $\text{p}_\text{i}\text{x}_\text{i}$ $\text{p}_\text{i}\text{X}_\text{i}^2$
$0$ $\frac{20}{42}$ $0$ $0$
$1$ $\frac{20}{42}$ $\frac{20}{42}$ $\frac{20}{42}$
$2$ $\frac{2}{42}$ $\frac{4}{42}$ $\frac{8}{42}$
    $\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{4}{7}$ $\sum\text{p}_\text{i}\text{x}_\text{i}^2=\frac{2}{3}$
Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{4}{7}$
Variance $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-\big(\sum\text{p}_\text{i}\text{x}_\text{i}\big)^2$
$=\frac{2}{3}-\Big(\frac{4}{7}\Big)^2=\frac{50}{147}$
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Question 155 Marks
From a lot containing 25 items, 5 of which are defective, 4 are choosen at random. Let X be the number of defective found. Obtain the probability distribution of X if the item are chosen without replacement.
Answer
Let X denote the number of defective item in a sample of 4 items drawn from a bag containing 5 defective items and 20 good items. Then, X can take values 0, 1, 2, 3 and 4.
Now,
P(X = 0)
= P(no defective item)
$=\frac{\text{}^{20}\text{C}_4}{\text{}^{25}\text{C}_4}$
$=\frac{4845}{12650}$
$=\frac{969}{2530}$
P(X = 1)
= P(1 defective item)
$=\frac{\text{}^5\text{C}_1\times\text{}^{20}\text{C}_3}{\text{}^{25}\text{C}_4}$
$=\frac{5700}{12650}$
$=\frac{114}{253}$
P(X = 2)
= P(2 defective item)
$=\frac{\text{}^5\text{C}_2\times\text{}^{20}\text{C}_2}{\text{}^{25}\text{C}_4}$
$=\frac{1900}{12650}$
$=\frac{38}{253}$
P(X = 3)
= P(3 defective item)
$=\frac{\text{}^5\text{C}_3\times\text{}^{20}\text{C}_1}{\text{}^{25}\text{C}_4}$
$=\frac{200}{12650}$
$=\frac{4}{253}$
P(X = 4)
= P(4 defective item)
$=\frac{\text{}^5\text{C}_4}{\text{}^{25}\text{C}_4}$
$=\frac{5}{12650}$
$=\frac{1}{2530}$
Thus, the probability distribution of X is given by
$\text{X}$
$0$
$1$
$2$
$3$
$4$
$\text{P}(\text{X})$
$\frac{969}{2530}$
$\frac{114}{253}$
$\frac{38}{253}$
$\frac{4}{253}$
$\frac{1}{2530}$
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Question 165 Marks
Two bad eggs are accidently mixed up with ten good ones. Three eggs are drawn at random with replacement from this lot. Compute the mean for the number of bad eggs drawn.
Answer
Let X denote the number of bad eggs in a sample of 3 eggs drawn from a lot containing 2 bad eggs and 10 good eggs. Then, X can take the values 0, 1 and 2. P(X = 0) = P(no bad egg) $=\frac{\text{}^{10}\text{C}_3}{\text{}^{12}\text{C}_3}$ $=\frac{120}{220}$ $=\frac{6}{11}$ P(X = 1) = P(1 bad egg) $=\frac{\text{}^{2}\text{C}_1\times\text{}^{10}\text{C}_2}{\text{}^{12}\text{C}_3}$ $=\frac{90}{220}$ $=\frac{9}{22}$ P(X = 2) = P(2 bad eggs) $=\frac{\text{}^{2}\text{C}_2\times\text{}^{10}\text{C}_1}{\text{}^{12}\text{C}_3}$ $=\frac{10}{220}$ $=\frac{1}{22}$ Thus, the probability distribution of X is given by
$\text{X}$ $\text{P}(\text{X})$
$0$ $\frac{6}{11}$
$1$ $\frac{9}{22}$
$2$ $\frac{1}{22}$
Computation of mean
$\text{x}_\text{i}$ $\text{p}_\text{i}$ $\text{x}_\text{i}\text{p}_\text{i}$
$0$ $\frac{6}{11}$ $0$
$1$ $\frac{9}{22}$ $\frac{9}{22}$
$2$ $\frac{1}{22}$ $\frac{1}{11}$
    $\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{1}{2}$
Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{1}{2}$
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Question 175 Marks
Find the mean, variance and standard deviation of the number of tails in three tosses of a coin.
Answer
Let X denotes the number of tails in three tosses of a coin. Then, X can take the values 0, 1, 2 and 3. P(X = 0) = P(HHH) $=\frac{1}{8},$ P(X = 1) = P(THH or HHT or HTH) $=\frac{3}{8}$ P(X = 2) = P(TTH or THT or HTT) $=\frac{3}{8},$ P(X = 3) = P(TTT) $=\frac{1}{8}$ Thus, the probability distribution of X is given by
$0$ $\frac{1}{8}$
$1$ $\frac{3}{8}$
$2$ $\frac{3}{8}$
$3$ $\frac{3}{8}$
Computation of mean and step deviation
$\text{x}_\text{i}$ $\text{p}_\text{i}$ $\text{p}_\text{i}\text{x}_\text{i}$ $\text{p}_\text{i}\text{x}_\text{i}^2$
$0$ $\frac{1}{8}$ $0$ $0$
$1$ $\frac{3}{8}$ $\frac{3}{8}$ $\frac{32}{221}$
$2$ $\frac{3}{8}$ $\frac{6}{8}$ $\frac{4}{221}$
$3$ $\frac{1}{8}$ $\frac{3}{8}$  
    $\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{3}{2}$ $\sum\text{p}_\text{i}\text{x}_\text{i}^2=3$
Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{3}{2}$ Mean $=\frac{34}{221}$ Variance $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-(\text{Mean})^2$ $3-\Big(\frac{3}{2}\Big)^2$ $=\frac{3}{4}$Step Deviation $=\sqrt{\text{Variance}}$
$=\sqrt{\frac{3}{4}}$
$=0.87$
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Question 185 Marks
Three cards are cdrawn successively with replacement from a well shffled deck of 52 cards. A random variable X denotes the number of hearts in the three cards drawn. Determine the probability distribution of X.
Answer
Three cards are thrown with replacement. Let X denote the numbers of hearts if three cards are drawn.
So, X has values 0, 1, 2, 3
$\text{P}(\text{X}=0)=\text{P}\big(\overline{\text{H}}_1\big)\times\text{P}\big(\overline{\text{H}}_2\big)\times\text{P}\big(\overline{\text{H}}_3\big)$
$=\frac{39}{52}\times\frac{39}{52}\times\frac{39}{52}$
$=\frac{27}{26}$
$\text{P}(\text{X}=1)=\text{P}(\text{H}_1)\text{P}\big(\overline{\text{H}}_2\big)\text{P}\big(\overline{\text{H}}_3\big)+\text{P}\big(\overline{\text{H}}_1\big)\text{P}(\text{H}_2)\text{P}\big(\overline{\text{H}}_3\big)+\text{P}\big(\overline{\text{H}}_1\big)\text{P}\big(\overline{\text{H}}_2\big)\text{P}(\text{H}_3)$
$=\frac{13}{52}\times\frac{39}{52}\times\frac{39}{52}+\frac{39}{52}\times\frac{13}{52}\times\frac{39}{52}+\frac{39}{52}\times\frac{13}{52}$
$=\frac{27}{26}$
$\text{P}(\text{X}=2)=\text{P}(\text{H}_1)\text{P}(\text{H}_2)\text{P}\big(\overline{\text{H}}_3\big)+\text{P}(\text{H}_1)\text{P}\big(\overline{\text{H}}_2\big)\text{P}(\text{H}_3)+\text{P}\big(\overline{\text{H}}_1\big)\text{P}(\text{H}_2)\text{P}(\text{H}_3)$
$=\frac{13}{52}\times\frac{13}{52}\times\frac{39}{52}+\frac{13}{52}\times\frac{39}{52}\times\frac{13}{52}+\frac{13}{52}\times\frac{13}{52}\times\frac{13}{52}$
$=\frac{9}{64}$
$\text{P}(\text{X}=3)=\text{P}(\text{H}_1)\text{P}(\text{H}_2)\text{P}(\text{H}_3)$
$=\frac{13}{52}\times\frac{13}{52}\times\frac{13}{52}$
$=\frac{1}{64}$
So,
Required probability distribution is
$\text{X}:$
$0$
$1$
$2$
$3$
$\text{P}(\text{X}):$
$\frac{27}{64}$
$\frac{27}{64}$
$\frac{9}{64}$
$\frac{1}{64}$
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Question 195 Marks
A random variable X takes the values 0, 1, 2 and 3 such that:
P(X = 0) = P(X > 0) = P(X < 0); P(X = -3) = P(X = -2) = P(X = -1); P(X = 1) = P(X = 2) = P(X = 3).
Obtain the probability distribution of X.
Answer
Let P(X = 0) = k. Then,
P(X = 0) = P(X > 0) = P(X < 0)
⇒ P(X > 0) = k
P(X < 0) = k
$\therefore$ P(X = 0) + P(X > 0) + P(X < 0) = 1
⇒ k + k + k = 1
$\Rightarrow\text{k}=\frac{1}{3}$
Now,
P(X < 0) = k
⇒ P(X = -1) + P(X = -2) + P(X = -3) = k
⇒ 3P(X = -1) = k $[\because$ P(X = -1) = P(X = -2) = P(X = -3)$]$
$\Rightarrow\text{P}(\text{X}-1)=\frac{\text{k}}{3}$
$\Rightarrow\text{P}(\text{X}=-1)=\frac{1}{3}\times\frac{1}{3}=\frac{1}{9}$
$\therefore\ \text{P}(\text{X}=-1)=\text{P}(\text{X}=-2)=\text{P}(\text{X}=-3)=\frac{1}{9}$
Similarly,
P(X > 0) = k
$\Rightarrow\text{P}(\text{X}=1)=\text{P}(\text{X}=2)=\text{P}(\text{X}=3)=\frac{1}{9}$
Thus, the probability distribution is given by
$\text{X}_\text{i}$ $\text{P}_\text{i}$
$-3$ $\frac{1}{9}$
$-2$ $\frac{1}{9}$
$-1$ $\frac{1}{9}$
$1$ $\frac{1}{9}$
$2$ $\frac{1}{9}$
$3$ $\frac{1}{9}$
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Question 205 Marks
Three cards are drawn at random (without replacement) from a well shuffled pack of 52 cards. Find the probability distribution of number of red cards. Hence, find the mean of the distribution.
Answer
Let 'X' be the random variable which can assume values from 0 to 3.
P(X = 0)
$=\frac{\text{}^{26}\text{C}_3}{\text{}^{52}\text{C}_3}=\frac{2600}{22100}=\frac{2}{17}$
P(X = 1)
$=\frac{\text{}^{26}\text{C}_1\times\text{}^{26}\text{C}_2}{\text{}^{52}\text{C}_3}=\frac{8450}{22100}=\frac{13}{17}$
P(X = 2)
$=\frac{\text{}^{26}\text{C}_2\times\text{}^{26}\text{C}_1}{\text{}^{52}\text{C}_3}=\frac{8450}{22100}=\frac{13}{17}$
P(X = 3)
$=\frac{\text{}^{26}\text{C}_3}{\text{}^{52}\text{C}_3}=\frac{2600}{22100}=\frac{2}{17}$
Probability distribution of X:
$\text{X}=\text{x}_\text{i}$
$0$
$1$
$2$
$3$
$\text{P}(\text{X} =\text{x}_\text{i})$
$\frac{2}{17}$
$\frac{13}{34}$
$\frac{13}{34}$
$\frac{2}{17}$
$=\sum\limits_{\text{i=0}}^3(\text{x}_\text{i}\times\text{p}_\text{i})$
$=\text{x}_0\text{p}_0+\text{x}_1\text{p}_1+\text{x}_2\text{p}_2+\text{x}_3\text{p}_3$
$=0\times\frac{2}{17}+1\times\frac{13}{34}+2\times\frac{13}{34}+3\times\frac{2}{17}$
$=\frac{13+26+12}{34}$
$=\frac{51}{34}$
$=\frac{3}{2}$
$=1.5$
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Question 215 Marks
Find the mean and variance of the number of tails in three tosses of a coin.
Answer
We know that in a throw of a coin. $\text{P}(\text{H})=\frac{1}{2},\text{P}(\text{T})=\frac{1}{2}$ Let X denote the number of heads in three tosses of a coin. So, X = 0, 1, 2, 3 P(X = 0) = P(T)P(T)P(T) $=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$ $=\frac{1}{8}$ P(X = 1) = P(H)P(T)P(T) + P(T)P(H)P(T) + P(T)P(T)P(H) $=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$ $=\frac{3}{8}$ P(X = 2) = P(H)P(H)P(T) + P(H)P(T)P(H) + P(T)P(H)P(H) $=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$ $=\frac{3}{8}$ P(X = 3) = P(H)P(H)P(H) $=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$ $=\frac{1}{8}$ So,
$\text{x}_\text{i}$ $\text{p}_\text{i}$ $\text{x}_\text{i}\text{p}_\text{i}$ $\text{x}_\text{i}^2\text{p}_\text{i}$
$0$ $\frac{1}{8}$ $0$ $0$
$1$ $\frac{3}{8}$ $\frac{3}{8}$ $\frac{3}{8}$
$2$ $\frac{3}{8}$ $\frac{6}{8}$ $\frac{12}{8}$
$3$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{9}{8}$
    $\sum\text{xp}=\frac{3}{2}$ $\sum\text{x}^2\text{p}$
Mean $=\sum\text{xp}=\frac{3}{2}$ Variance $=\sum\text{x}^2\text{p}-(\text{Mean})^2$ $=3-\frac{9}{4}=\frac{3}{4}$
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Question 225 Marks
A pair of fair dice is thrown. Let X be the random variable which denotes the minimum of the two numbers which appear. Find the probability distribution, mean and variance of X.
Answer
A pair of fair dice is thrown. And X denote minimum of the two number appeared. So, X can values 2, 3, 4, 5, 6. $\text{P}(\text{X}=1)=\frac{11}{36}$ [Possible pairs: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)] $\text{P}(\text{X}=2)=\frac{9}{36}$ [Possible pairs: (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (4, 2), (5, 2), (6, 2)] $\text{P}(\text{X}=3)=\frac{7}{36}$ [Possible pairs: (3, 3), (3, 4), (3, 5), (3, 6), (4, 3), (5, 3), (6, 3)] $\text{P}(\text{X}=4)=\frac{5}{36}$ [Possible pairs: (4, 4), (4, 5), (4, 6), (5, 4), (6, 4)] $\text{P}(\text{X}=5)=\frac{3}{36}$ [Possible pairs: (4, 4), (5, 5), (5, 6)5, (6, 5)] $\text{P}(\text{X}=6)=\frac{1}{36}$ [Possible pairs: (6, 6)]
$\text{x}_\text{i}$ $\text{p}_\text{i}$ $\text{x}_\text{i}\text{p}_\text{i}$ $\text{x}_\text{i}^2\text{p}_\text{i}$
$1$ $\frac{11}{36}$ $\frac{11}{36}$ $\frac{11}{36}$
$2$ $\frac{9}{36}$ $\frac{18}{36}$ $\frac{36}{36}$
$3$ $\frac{7}{36}$ $\frac{21}{36}$ $\frac{63}{36}$
$4$ $\frac{5}{36}$ $\frac{20}{36}$ $\frac{80}{36}$
$5$ $\frac{3}{36}$ $\frac{15}{36}$ $\frac{75}{36}$
$6$ $\frac{1}{36}$ $\frac{6}{36}$ $\frac{36}{36}$
    $\sum\text{xp}=\frac{91}{36}$ $\sum\text{x}^2\text{p}=\frac{301}{36}$
Mean $=\sum\text{xp}$ Mean $=\frac{91}{36}$ Variance $=\sum\text{x}^2\text{p}-(\text{mean})^2$ $=\frac{301}{36}-\Big(\frac{91}{36}\Big)^2$ $=\frac{10836-8281}{1296}$ $=\frac{2555}{1296}$ Variance = 1.97 Probability distribution is
$\text{x}:$
$1$
$2$
$3$
$4$
$5$
$6$
$\text{p}(\text{x}):$
$\frac{11}{36}$
$\frac{9}{36}$
$\frac{7}{36}$
$\frac{5}{36}$
$\frac{3}{36}$
$\frac{1}{36}$
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Question 235 Marks
A class has 15 students whose ages are 14, 17, 15, 14, 21, 19, 20, 16, 18, 17, 20, 17, 16, 19 and 20 years respectively. One student is selected in such a manner that each has the same chance to being selected and the age X of the selected student is recorded. What is the probability distribution of the random variable X.
Answer
Here, X denote the number of two number or two dice thrown together.
So, X = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
So,
$\text{P}(\text{X}=2)=\frac{1}{36}$ [Possible pairs: (1, 1)]
$\text{P}(\text{X}=3)=\frac{2}{36}=\frac{1}{18}$ [Possible pairs: (1, 2), (2,1)]
$\text{P}(\text{X}=4)=\frac{3}{36}=\frac{1}{12}$ [Possible pairs: (1, 3), (2,2), (3, 1)]
$\text{P}(\text{X}=5)=\frac{4}{36}=\frac{1}{9}$ [Possible pairs: (1, 4), (2, 3), (3, 2), (4, 1)]
$\text{P}(\text{X}=6)=\frac{5}{36}$ [Possible pairs: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1)]
$\text{P}(\text{X}=7)=\frac{6}{36}=\frac{1}{6}$ [Possible pairs: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)]
$\text{P}(\text{X}=8)=\frac{5}{36}$ [Possible pairs: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)]
$\text{P}(\text{X}=9)=\frac{4}{36}=\frac{1}{9}$ [Possible pairs: (3, 6), (4, 5), (5, 4), (6, 3)]
$\text{P}(\text{X}=10)=\frac{3}{36}=\frac{1}{12}$ [Possible pairs: (4, 6), (5, 5), (6, 4)]
$\text{P}(\text{X}=11)=\frac{2}{36}=\frac{1}{18}$ [Possible pairs: (5, 6), (6,5)]
$\text{P}(\text{X}=12)=\frac{1}{36}$ [Possible pairs: (6, 6)]
So, required possibility distribution is
$\text{X}:$
$2$
$3$
$4$
$5$
$6$
$7$
$8$
$9$
$10$
$11$
$12$
$\text{P}(\text{X}):$
$\frac{1}{36}$
$\frac{1}{18}$
$\frac{1}{12}$
$\frac{1}{9}$
$\frac{5}{36}$
$\frac{1}{6}$
$\frac{5}{36}$
$\frac{1}{9}$
$\frac{1}{12}$
$\frac{1}{18}$
$\frac{1}{36}$
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Question 245 Marks
Two dice are drawn together and the number appearing on them noted. X denotes the sum of the two numbers. Assuming that the 36 outcomes are equally likely, what is the probability distribution of X?
Answer
Let X denote numbers on two die. Then, X can take the values 2, 3, 4, 5, 6, 7 , 8, 9, 10, and 12.
Sample space:
{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Now,
$\text{P}(\text{x}=2)=\frac{1}{36}$
$\text{P}(\text{x}=3)=\frac{2}{36}$
$\text{P}(\text{x}=4)=\frac{3}{36}$
$\text{P}(\text{x}=5)=\frac{4}{36}$
$\text{P}(\text{x}=6)=\frac{5}{36}$
$\text{P}(\text{x}=7)=\frac{6}{36}$
$\text{P}(\text{x}=8)=\frac{5}{36}$
$\text{P}(\text{x}=9)=\frac{4}{36}$
$\text{P}(\text{x}=10)=\frac{3}{36}$
$\text{P}(\text{x}=11)=\frac{2}{36}$
$\text{X}$
$2$
$3$
$4$
$5$
$6$
$7$
$8$
$9$
$10$
$11$
$12$
$\text{P}(\text{X})$
$\frac{1}{36}$
$\frac{2}{36}$
$\frac{3}{36}$
$\frac{4}{36}$
$\frac{5}{36}$
$\frac{6}{36}$
$\frac{5}{36}$
$\frac{4}{36}$
$\frac{3}{36}$
$\frac{2}{36}$
$\frac{1}{36}$
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Question 255 Marks
Two cards are drawn successively with replacement from well shuffled pack of 52 cards. Find the probability distribution of the number of kings.
Answer
Let X denote the number of kings in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
P(X = 0)
= P(no kings)
$=\frac{48}{52}\times\frac{48}{52}$
$=\frac{12\times12}{13\times13}$
$=\frac{144}{169}$
P(X = 1)
= P(1 king)
$=\frac{4}{52}\times\frac{48}{52}$
$=\frac{2\times12}{13\times13}$
$=\frac{24}{169}$
P(X = 2)
= P(2 kings)
$=\frac{4}{52}\times\frac{4}{52}$
$=\frac{1\times1}{13\times13}$
$=\frac{1}{169}$
Thus, the probability distribution of X is given by
$\text{X}$
$0$
$1$
$2$
$\text{P}(\text{X})$
$\frac{144}{169}$
$\frac{24}{169}$
$\frac{1}{169}$
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Question 265 Marks
Four balls are to be drawn without replacement from a box containing 8 red and 4 white balls. If X denotes the number of red balls drawn, then find the probability distribution of X.
Answer
A, four balls are to be drawn without replacement and X denote the number of red balls drawn.
So, X is a random variable that can take values 0, 1, 2, 3 or 4.
Now,
P(X = 0) = P(All white balls) = P(WWWW) $=\frac{4}{12}\times\frac{3}{11}\times\frac{2}{10}\times\frac{1}{9}=\frac{1}{495},$
P(X = 1) = P(One of red balls and three white balls) = P(RWWW) + P(WRWW) + P(WWRW) + P(WWWR)
$=\frac{8}{12}\times\frac{4}{11}\times\frac{3}{10}\times\frac{2}{9}+\frac{4}{12}\times\frac{8}{11}\times\frac{3}{10}\times\frac{2}{9}+\frac{4}{12}\times\frac{3}{11}\times\frac{8}{10}\times\frac{2}{9}+\frac{4}{12}\times\frac{3}{11}\times\frac{2}{10}\times\frac{8}{9}$
$=4\times\frac{8}{495}=\frac{32}{495}.$
P(X = 2) = P(Two red balls and two white balls) = P(RRWW) + P(RWRW) + P(RWWR) + P(WWRR) + P(W
P(X = 3) = P(Three red balls and one white ball) = P(RRRW) + P(RRWR) + P(RWRR) + P(WRRR)
$=\frac{8}{12}\times\frac{7}{11}\times\frac{6}{10}\times\frac{4}{9}+\frac{8}{12}\times\frac{7}{11}\times\frac{4}{10}\times\frac{6}{9}+\frac{8}{12}\times\frac{4}{11}\times\frac{7}{10}\times\frac{6}{9}+\frac{4}{12}\times\frac{8}{11}\times\frac{7}{10}\times\frac{6}{9}$
$=4\times\frac{56}{495}=\frac{224 }{495}.$
P(X = 4) = P(All red balls) = P(RRRR) $=\frac{8}{12}\times\frac{7}{11}\times\frac{6}{10}\times\frac{5}{9}$
$=\frac{70}{495}$
So, the probability distribution of X is as follows:
$\text{X}$
$0$
$1$
$2$
$3$
$4$
$\text{P}(\text{X})$
$\frac{1}{495}$
$\frac{32}{495}$
$\frac{168}{495}$
$\frac{224}{495}$
$\frac{70}{495}$
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Question 275 Marks
Write the values of 'a' for which the following distribution of probabilities becomes a probability distributioin:
$\text{X}=\text{x}_\text{i}:$ $-2$ $-1$ $0$ $1$
$\text{P}(\text{X}=\text{x}_\text{i}):$ $\frac{1-\text{a}}{4}$ $\frac{1+2\text{a}}{4}$ $\frac{1-2\text{a}}{4}$ $\frac{1+\text{a}}{4}$
Answer
Here,
$\text{X}:$ $-2$ $-1$ $0$ $1$
$\text{P}(\text{X}):$ $\frac{1-\text{a}}{4}$ $\frac{1+2\text{a}}{4}$ $\frac{1-2\text{a}}{4}$ $\frac{1+\text{a}}{4}$
Now, $\sum\text{P}(\text{X})=\text{P}(-)+\text{P}(-1)+\text{P}(0)+\text{P}(1)$
$=\frac{1-\text{a}}{4}+\frac{1+2\text{a}}{4}+\frac{1-2\text{a}}{4}+\frac{1+\text{a}}{4}$
$=\frac{1-\text{a}-1+2\text{a}+1-2\text{a}+1+\text{a}}{4}$
$=1$
So, the sum of the probability must be positive and less than or equal to 1.
Now,
$0\leq\frac{1-\text{a}}{4}\leq1\Rightarrow0\leq1-\text{a}\leq4\Rightarrow-1\leq-\text{a}\leq3\Rightarrow1\geq\text{a}\geq-3$
$0\leq\frac{1+2\text{a}}{4}\leq1\Rightarrow0\leq1+2\text{a}\leq4\Rightarrow-1\leq2\text{a}\leq3\Rightarrow-\frac{1}{2}\leq\text{a}\leq\frac{3}{2}$
$0\leq\frac{1+\text{a}}{4}\leq1\Rightarrow0\leq1+\text{a}\leq4\Rightarrow-1\leq\text{a}\leq3$
Therefore, $-\frac{1}{2}\leq\text{a}\leq\frac{1}{2}$
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5 Marks Questions - MATHS STD 12 Science Questions - Vidyadip