Probability Distributions of a Random Variable

Q 1 :

If the mean of the following probability distribution of a random variable X :

X 0 2 4 6 8

P(X) $a$ $2 a$ $a + b$ $2 b$ $3 b$

is $\frac{46}{9},$ then the variance of the distribution is [2024]

$\frac{151}{27}$
$\frac{173}{27}$
$\frac{566}{81}$
$\frac{581}{81}$

1

2

3

4

(3)

X	0	2	4	6	8
P(X)	$a$	$2 a$	$a + b^{3}$	$2 b$	$3 b$

$Mean = \sum x_{i} P (x_{i})$

$\Rightarrow \frac{46}{9} = 0 + 4 a + 4 a + 4 b + 12 b + 24 b \Rightarrow \frac{46}{9} = 8 a + 40 b$

$\Rightarrow 36 a + 180 b = 23$ ...(i)

Also, $\sum_{i - 1}^{n} P_{i} = 1$

$\Rightarrow 4 a + 6 b = 1$ ...(ii)

On solving (i) and (ii), we get

$a = \frac{1}{12}, b = \frac{1}{9}$

Now, $σ^{2} = \sum x_{i}^{2} P (x_{i}) - (\sum x_{i} P {(x_{i}))}^{2}$

$= 0 + 4 \times 2 a + 16 (a + b) + 36 (2 b) + 64 (3 b) - {(\frac{46}{9})}^{2}$

$= 8 (a + 2 (a + b) + 9 b + 24 b) - {(\frac{46}{9})}^{2}$

$= 8 (3 a + 35 b) - {(\frac{46}{9})}^{2}$ $= 8 (\frac{3}{12} + \frac{35}{9}) - {(\frac{46}{9})}^{2}$

$= 8 (\frac{149}{36}) - {(\frac{46}{9})}^{2}$ $= \frac{566}{81}$

Q 2 :

Three rotten apples are accidentally mixed with fifteen good apples. Assuming the random variable $x$ to be the number of rotten apples in a draw of two apples, the variance of $x$ is [2024]

$\frac{37}{153}$
$\frac{57}{153}$
$\frac{40}{153}$
$\frac{47}{153}$

1

2

3

4

(3)

We have,

Number of rotten apples = 3

Number of good apples = 15

$x$ is the number of rotten apples

$x$	0	1	2
$p (x)$	$\frac{{}^{15}{C_{2}}}{{}^{18}{C_{2}}} = \frac{105}{153}$	$\frac{{}^{3}{C_{1} \times {}^{15}C_{1}}}{{}^{18}{C_{2}}}$ $= \frac{3 \times 15}{153} = \frac{45}{153}$	$\frac{{}^{3}{C_{2}}}{{}^{18}{C_{2}}} = \frac{3}{153}$

$E (x) = 0 \times \frac{105}{153} + 1 \times \frac{45}{153} + \frac{6}{153} = \frac{51}{153} = \frac{1}{3}$

$E (x^{2}) = 0 \times \frac{105}{153} + 1 \times \frac{45}{153} + \frac{12}{153} = \frac{57}{153}$

$∴$ $Var (x) = E (x^{2}) - {(E (x))}^{2} = \frac{57}{153} - {(\frac{1}{3})}^{2}$ $= \frac{57}{153} - \frac{1}{9} = \frac{40}{153}$

Q 3 :

From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. If the variance of $X$ is $σ^{2},$ then $96 σ^{2}$ is equal to _____________ . [2024]

(56)

Total number of items = 10

Number of defective items = 3

$∴$ Number of non-defective items = 7

X	0	1	2	3
P(X)	$\frac{{}^{7}{C_{5}}}{{}^{10}{C_{5}}} = \frac{1}{12}$	$\frac{{}^{7}{C_{4} \times {}^{3}{C_{1}}}}{{}^{10}{C_{5}}} = \frac{5}{12}$	$\frac{{}^{7}{C_{3} \times {}^{3}{C_{2}}}}{{}^{10}{C_{5}}} = \frac{5}{12}$	$\frac{{}^{7}{C_{2} \times {}^{3}{C_{3}}}}{{}^{10}{C_{5}}} = \frac{1}{12}$

Now, $σ^{2} = \sum {(X - \bar{X})}^{2} P (X)$

$\bar{X} = \sum X P (X) = 0 + \frac{5}{12} + \frac{10}{12} + \frac{3}{12} = \frac{18}{12} = \frac{3}{2}$

$∴$ $σ^{2} = \sum {(X - \frac{3}{2})}^{2} P (X)$

$= \frac{9}{4} \times \frac{1}{12} + \frac{1}{4} \times \frac{5}{12} + \frac{1}{4} \times \frac{5}{12} + \frac{9}{4} \times \frac{1}{12} = \frac{7}{12}$

$\Rightarrow 96 σ^{2} = 8 \times 7 = 56$

Q 4 :

From a lot of 12 items containing 3 defectives, a sample of 5 items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If the variance of $X$ is $\frac{m}{n},$ where $g c d (m, n) = 1,$ then $n - m$ is equal to ______ . [2024]

(71)

X	0	1	2	3
P(X)	$\frac{{}^{9}{C_{5}}}{{}^{12}{C_{5}}}$	$\frac{{}^{9}{C_{4} \times {}^{3}{C_{1}}}}{{}^{12}{C_{5}}}$	$\frac{{}^{9}{C_{3} \times {}^{3}{C_{2}}}}{{}^{12}{C_{5}}}$	$\frac{{}^{9}{C_{2} \times {}^{3}{C_{3}}}}{{}^{12}{C_{5}}}$

$Mean = \sum X P (X) = 0 + \frac{126 \times 3}{792} + \frac{2 \times 84 \times 3}{792} + \frac{3 \times 36 \times 1}{792}$

$= \frac{378 + 504 + 108}{792} = \frac{990}{792} = \frac{55}{44}$

$Variance = \sum X^{2} P (X) - (\sum X P {(X))}^{2}$

$= \frac{95}{44} - {(\frac{55}{44})}^{2} = \frac{4180 - 3025}{1936} = \frac{1155}{1936} = \frac{105}{176} = \frac{m}{n}$

$So, n - m = 176 - 105 = 71$

Q 5 :

Three balls are drawn at random from a bag containing 5 blue and 4 yellow balls. Let the random variables $X$ and $Y$ respectively denote the number of blue and yellow balls. If $\bar{X}$ and $\bar{Y}$ are the means of $X$ and $Y$ respectively, then $7 \bar{X} + 4 \bar{Y}$ is equal to ________ . [2024]

(17)

Blue balls (X)	0	1	2	3
Probability P(X)	$\frac{{}^{5}{C_{0} {}^{4}{C_{3}}}}{{}^{9}{C_{3}}}$	$\frac{{}^{5}C_{1} \times {}^{4}{C_{2}}}{{}^{9}{C_{3}}}$	$\frac{{}^{5}{C_{2} \times {}^{4}{C_{1}}}}{{}^{9}{C_{3}}}$	$\frac{{}^{5}{C_{3} \times {}^{4}{C_{0}}}}{{}^{9}{C_{3}}}$

$\bar{X} = \frac{0 + 5 \times 6 + 2 \times 10 \times 4 + 3 \times 10 \times 1}{84}$

$\Rightarrow \bar{X} = \frac{140}{84} \Rightarrow 7 \bar{X} = \frac{140}{12} = \frac{35}{3}$

Yellow balls (X)	0	1	2	3
Probability P(Y)	$\frac{{}^{5}{C_{3} \times {}^{4}{C_{0}}}}{{}^{9}{C_{3}}}$	$\frac{{}^{5}{C_{2} \times {}^{4}{C_{1}}}}{{}^{9}{C_{3}}}$	$\frac{{}^{5}{C_{1} \times {}^{4}{C_{2}}}}{{}^{9}{C_{3}}}$	$\frac{{}^{5}{C_{0} \times {}^{4}{C_{3}}}}{{}^{9}{C_{3}}}$

$\bar{Y} = \frac{0 + 10 \times 4 + 2 \times 5 \times 6 + 3 \times 4}{84} = \frac{112}{84} \Rightarrow 4 \bar{Y} = \frac{16}{3}$

$7 \bar{X} + 4 \bar{Y} = \frac{35}{3} + \frac{16}{3} = 17$

Q 6 :

A fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required and let $a = P (X = 3), b = P (X \geq 3)$ and $c = P (X \geq 6 ∣ X > 3) .$ Then $\frac{b + c}{a}$ is equal to _______ . [2024]

(12)

Probability of getting six $= \frac{1}{6}$

Probability of not getting a six $= \frac{5}{6}$

Let $X$ denote the number of tosses required. Then,

$a = P (X = 3) = \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} = \frac{5^{2}}{6^{3}} = \frac{25}{216}$

$b = P (X \geq 3) = 1 - [P (X = 1) + P (X = 2)]$ $= 1 - [\frac{1}{6} + \frac{5}{6} \times \frac{1}{6}] = \frac{25}{36}$

$c = P (X \geq 6 ∣ X > 3) = \frac{P [(X \geq 6) \cap (X > 3)]}{P (X > 3)} = \frac{P (X \geq 6)}{P (X > 3)}$

$= \frac{{(\frac{5}{6})}^{5} \times \frac{1}{6} + {(\frac{5}{6})}^{6} \times \frac{1}{6} + \dots}{1 - \frac{1}{6} - (\frac{5}{6}) \times (\frac{1}{6}) - {(\frac{5}{6})}^{2} \times (\frac{1}{6})}$

$= \frac{{(\frac{5}{6})}^{5} \times \frac{1}{6} [1 + \frac{5}{6} + {(\frac{5}{6})}^{2} + \dots]}{\frac{25}{36} - \frac{25}{216}}$ $= \frac{25}{36}$

$∴$ $\frac{b + c}{a} = \frac{\frac{25}{36} + \frac{25}{36}}{\frac{25}{216}} = 12$

Q 7 :

If the probability that the random variable X takes the value x is given by $P (X = x) = k (x + 1) 3^{- x}$ , x = 0, 1, 2, 3, ..., where k is a constant, then $P (X \geq 3)$ , is equal to [2025]

$\frac{8}{27}$
$\frac{4}{9}$
$\frac{1}{9}$
$\frac{7}{27}$

1

2

3

4

(3)

Given, $P (X = x) = k (x + 1) 3^{- x}$

We have, $\sum_{x = 0}^{\infty} k (x + 1) 3^{- x} = 1$

$\Rightarrow \frac{1}{k} = 1 + \frac{2}{3} + \frac{3}{3^{2}} + \frac{4}{3^{3}} + . . .$ ... (i)

$\Rightarrow \frac{1}{3 k} = \frac{1}{3} + \frac{2}{3^{2}} + \frac{3}{3^{3}} + . . .$ ... (ii)

Subtracting equation (ii) from (i), we get

$\frac{1}{k} - \frac{1}{3 k} = 1 + \frac{1}{3} + \frac{1}{3^{2}} + . . .$

$\Rightarrow \frac{2}{3 k} = \frac{1}{1 - \frac{1}{3}} \Rightarrow k = \frac{4}{9}$

Now, $P (X \geq 3) = 1 - [P (X = 0) + P (X = 1) + P (X = 2)]$

$= 1 - k (1 + \frac{2}{3} + \frac{3}{9}) = 1 - \frac{4}{9} (2) = 1 - \frac{8}{9} = \frac{1}{9}$ .

Q 8 :

Let a random variable X take values 0, 1, 2, 3, with P(X = 0) = P(X = 1) = p, P(X = 2) = P(X = 3) and $E (X^{2}) = 2 E (X)$ . Then the value 8p – 1 is : [2025]

0
3
1
2

1

2

3

4

(4)

We have, P(X = 0) = P(X = 1) = p and let P(X = 2) = P(X = 3)= q

$E (X) = \sum_{i = 0}^{3} x_{i}^{} P (X = x_{i}) = 0 + 1 p + 2 q + 3 q = p + 5 q$

$E (X^{2}) = \sum_{i = 0}^{3} x_{i}^{2} P (X = x_{i})$

$= 0 \cdot p + 1 \cdot p + 4 \cdot q + 9 \cdot q = p + 13 q$

Now, $E (X^{2}) = 2 E (X)$ [Given]

$\Rightarrow p + 13 q = 2 (p + 5 q)$

$\Rightarrow p = 3 q$ ... (i)

Also, $\sum_{i = 0}^{3} P (X = x_{i}) = 1$

$\Rightarrow 2 p + 2 q = 1$

$\Rightarrow 2 (p + q) = 1$ ... (ii)

On solving (i) and (ii), we get

$q = \frac{1}{8} and p = \frac{3}{8}$

$∴ 8 p - 1 = 3 - 1 = 2$ .

Q 9 :

Let $A = [a_{i j}]$ be a $2 \times 2$ matrix such that $a_{i j} \in {0, 1}$ for all i and j. Let the random variable X denote the possible values of the determinant of the matrix A. Then, the variance of X is : [2025]

$\frac{5}{8}$
$\frac{3}{8}$
$\frac{1}{4}$
$\frac{3}{4}$

1

2

3

4

(2)

We have, $A = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}]$ , where $a_{i j} \in {0, 1}$

$\Rightarrow | A | = a_{11} a_{22} - a_{12} a_{21} = {- 1, 0, 1}$

Total outcomes = $2^{4}$ = 16, out of which 3 cases will be for value –1, 3 cases for 1, and rest 10 cases for 0.

X	–1	0	1
P(X)	3/16	10/16	3/16

Now, $\bar{X} = \sum_{}^{} (X) P (X) = - 1 (\frac{3}{16}) + 0 (\frac{10}{16}) + 1 (\frac{3}{16}) = 0$

$∴ Variance = \sum_{}^{} X_{i}^{2} P (X) - {[\sum_{}^{} X P (X)]}^{2}$

$= 1 \times \frac{3}{16} + 0 \times \frac{10}{16} + 1 \times \frac{3}{16} - 0 = \frac{3}{8}$ .

Q 10 :

A pair of dice is thrown 5 times. For each throw, a total of 5 is considered a success. If the probability of at least 4 successes is $\frac{k}{3^{11}}$ , then $k$ is equal to [2023]

82
164
75
123

1

2

3

4

(4)

For a total of 5 we have $(1, 4), (4, 1), (2, 3), (3, 2)$

$So, probability of success = \frac{4}{36} = \frac{1}{9} = p$

$and probability of failure = \frac{8}{9} = q$

Now, $P (X \geq 4) = {}^{5}{C_{4}} {(\frac{1}{9})}^{4} \cdot {(\frac{8}{9})}^{1} + {}^{5}{C_{5}} {(\frac{1}{9})}^{5}$

$\Rightarrow \frac{k}{3^{11}} = \frac{40}{9^{5}} + \frac{1}{9^{5}} \Rightarrow k = 41 \times 3 = 123$

Q 11 :

If the probability that the random variable X takes values $x$ is given by $P (X = x) = k (x + 1) 3^{- x}, x = 0, 1, 2, 3, \dots$ , where $k$ is a constant, then $P (X \geq 2)$ is equal to [2023]

$\frac{7}{27}$
$\frac{7}{18}$
$\frac{11}{18}$
$\frac{20}{27}$

1

2

3

4

(1)

$\sum_{x = 0}^{\infty} P (X = x) = 1 \Rightarrow k (1 + \frac{2}{3} + \frac{3}{3^{2}} + \frac{4}{3^{3}} + \dots \infty) = 1$

Let $S = 1 + \frac{2}{3} + \frac{3}{3^{2}} + \frac{4}{3^{3}} + \dots \infty \Rightarrow \frac{S}{3} = \frac{1}{3} + \frac{2}{3^{2}} + \frac{3}{3^{3}} + \dots \infty$

$\Rightarrow \frac{2 S}{3} = 1 + \frac{1}{3} + \frac{1}{3^{2}} + \dots \infty \Rightarrow \frac{2 S}{3} = \frac{1}{1 - \frac{1}{3}}$

$\Rightarrow \frac{2 S}{3} = \frac{3}{2} \Rightarrow S = \frac{9}{4} \Rightarrow k = \frac{4}{9}$

Now, $P (X \geq 2) = 1 - P (X = 0) - P (X = 1) = 1 - \frac{4}{9} (1 + \frac{2}{3}) = \frac{7}{27}$

Q 12 :

Let a die be rolled $n$ times. Let the probability of getting odd numbers seven times be equal to the probability of getting odd numbers nine times. If the probability of getting even numbers twice is $\frac{k}{2^{15}}$ , then $k$ is equal to [2023]

30
15
60
90

1

2

3

4

(3)

$Given, P (getting odd numbers 7 times) = P (getting odd numbers 9 times)$

$\Rightarrow {}^{n}{C_{7}} {(\frac{1}{2})}^{7} {(\frac{1}{2})}^{n - 7} = {}^{n}{C_{9}} {(\frac{1}{2})}^{9} {(\frac{1}{2})}^{n - 9}$

$\Rightarrow {}^{n}{C_{7}} = {}^{n}{C_{9}}$ $∴$ $n = 9 + 7 = 16$

$Now, P (getting even number twice) = {}^{16}C_{2} {(\frac{1}{2})}^{14} {(\frac{1}{2})}^{2} =$ $\frac{{}^{16}{C_{2}}}{2^{16}}$

$= \frac{16!}{2! \times 14!} \times \frac{1}{2^{16}} = \frac{16 \times 15}{2 \times 2^{16}} = \frac{15 \times 4}{2^{15}} = \frac{k}{2^{15}}$

$So, k = 60$

Q 13 :

Two dice A and B are rolled. Let the numbers obtained on A and B be $α$ and $β$ respectively. If the variance of $α - β$ is $\frac{p}{q}$ , where $p$ and $q$ are co-prime, then the sum of the positive divisors of $p$ is equal to [2023]

72
48
36
31

1

2

3

4

(2)

Cases	$α - β$	$P$
(6,1)	5	1/36
(6,2), (5,1)	4	2/36
(6,3), (5,2), (4,1)	3	3/36
(6,4), (5,3), (4,2), (3,1)	2	4/36
(6,5), (5,4), (4,3), (3,2), (2,1)	1	5/36
(6,6), (5,5), (4,4), (3,3), (2,2), (1,1)	0	6/36
(1,2), (2,3), (3,4), (4,5), (5,6)	−1	5/36
(1,3), (2,4), (3,5), (4,6)	−2	4/36
(1,4), (2,5), (3,6)	−3	3/36
(1,5), (2,6)	−4	2/36
(1,6)	−5	1/36

$E (x) = \sum x_{i} P (x_{i}) = 0$

$E (x^{2}) = \sum x_{i}^{2} P (x_{i}) = 2 [\frac{25}{36} + \frac{32}{36} + \frac{27}{36} + \frac{16}{36} + \frac{5}{36}]$

$= 2 [\frac{105}{36}] = \frac{35}{6}$

$Variance ρ^{2} = E (x^{2}) - {[E (x)]}^{2} = \frac{35}{6} = \frac{p}{q} \Rightarrow p = 35 = 7 \times 5$

$∴$ $Sum of divisors = (5^{0} + 5^{1}) (7^{0} + 7^{1}) = 6 \times 8 = 48$

Q 14 :

A coin is biased so that the head is 3 times as likely to occur as tail. This coin is tossed until a head or three tails occur. If $X$ denotes the number of tosses of the coin, then the mean of $X$ is [2023]

$\frac{21}{16}$
$\frac{37}{16}$
$\frac{81}{64}$
$\frac{15}{16}$

1

2

3

4

(1)

$Let x be the probability of getting tail.$

$\Rightarrow P (T) = x$ $∴$ $P (H) = 3 x .$

$We know that sum of all probabilities of events of an experiment = 1$

$∴$ $x + 3 x = 1 \Rightarrow x = \frac{1}{4}$ $\Rightarrow P (T) = \frac{1}{4}, P (H) = \frac{3}{4}$

X	1	2	3
P(X)	$\frac{3}{4}$	$\frac{1}{4} \times \frac{3}{4}$	${(\frac{1}{4})}^{3} + {(\frac{1}{4})}^{2} \times (\frac{3}{4})$

$Now, Mean \bar{X} = \sum x_{i} P (x_{i})$ $= 1 \times \frac{3}{4} + 2 \times \frac{1}{4} \times \frac{3}{4} + 3 (\frac{1}{64} + \frac{3}{64})$

$= \frac{3}{4} + \frac{6}{16} + \frac{3}{16} = \frac{21}{16}$

Q 15 :

The random variable $X$ follows binomial distribution $B (n, p)$ , for which the difference of the mean and the variance is 1. If $2 P (X = 2) = 3 P (X = 1),$ then $n^{2} P (X > 1)$ is equal to [2023]

15
11
16
12

1

2

3

4

(2)

$n p - n p q = 1$

$\Rightarrow n p (1 - q) = 1$

$\Rightarrow n p^{2} = 1$ $[∵ p = 1 - q]$ ...(i)

$Now, 2 P (X = 2) = 3 P (X = 1)$ ...(ii)

$\Rightarrow 2 \cdot {}^{n}C_{2} p^{2} q^{n - 2} = 3 \cdot {}^{n}C_{1} p q^{n - 1}$

$\Rightarrow n p - p = 3 q$ $[∵ q = 1 - p]$

$\Rightarrow n p + 2 p = 3$ $[from (i) and (ii)]$

$\Rightarrow p = \frac{1}{2}$

$From (i), n p^{2} = 1$

$\Rightarrow n \times {(\frac{1}{2})}^{2} = 1 \Rightarrow n = 4$

$P (X > 1) = 1 - P (X = 0) + P (X = 1)$

$= 1 - [4 C_{0} {(\frac{1}{2})}^{4} + 4 C_{1} (\frac{1}{2}) {(\frac{1}{2})}^{3}] = \frac{11}{16}$

$∴$ $n^{2} p (x > 1) = 4^{2} \times \frac{11}{16} = 11$

Q 16 :

In a binomial distribution $B (n, p)$ , the sum and the product of the mean and the variance are 5 and 6 respectively, then $6 (n + p - q)$ is equal to [2023]

50
53
52
51

1

2

3

4

(3)

Let mean of distribution = $x$

and variance of distribution = $y$

$∴$ $x + y = 5 and x y = 6$

$\Rightarrow x = 3 and y = 2$ $[∵ mean > variance]$

Now, mean of binomial distribution = $n p$

and variance of binomial distribution = $n p q$

$∴$ $x = n p and y = n p q$

$\Rightarrow n p = 3 and n p q = 2 \Rightarrow 3 \cdot q = 2 \Rightarrow q = \frac{2}{3}$

Now, we know $p + q = 1$

$\Rightarrow p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$ and $n = \frac{3}{p} = \frac{1}{1 / 3} = 9$

$∴$ $6 (n + p - q) = 6 (9 + \frac{1}{3} - \frac{2}{3}) = 6 (\frac{27 + 1 - 2}{3}) = 52$

Q 17 :

Three rotten apples are mixed accidently with seven good apples and four apples are drawn one by one without replacement. Let the random variable $X$ denote the number of rotten apples. If $μ$ and $σ^{2}$ represent mean and variance of $X$ , respectively, then $10 (μ^{2} + σ^{2})$ is equal to [2023]

250
20
25
30

1

2

3

4

(2)

Let random variable $X$ denote the number of rotten apples.

$X_{i}$	$P_{i}$
$X = 0$	$\frac{{}^{7}{C_{4}} \times {}^{3}C_{0}}{{}^{10}{C_{4}}} = \frac{1}{6}$
$X = 1$	$\frac{{}^{3}{C_{1}} \times {}^{7}C_{3}}{{}^{10}{C_{4}}} = \frac{1}{2}$
$X = 2$	$\frac{{}^{3}{C_{2}} \times {}^{7}C_{2}}{{}^{10}{C_{4}}} = \frac{3}{10}$
$X = 3$	$\frac{{}^{3}{C_{3}} \times {}^{7}C_{1}}{{}^{10}{C_{4}}} = \frac{1}{30}$

$Mean (μ) = \sum X_{i} P_{i}$

$= 0 \times \frac{1}{6} + 1 \times \frac{1}{2} + 2 \times \frac{3}{10} + 3 \times \frac{1}{30} = \frac{6}{5}$

$Variance (σ^{2}) = \sum X_{i}^{2} P_{i} - μ^{2}$

$= 0 \times \frac{1}{6} + 1 \times \frac{1}{2} + 4 \times \frac{3}{10} + 9 \times \frac{1}{30} - \frac{36}{25} = \frac{14}{25}$

$So, 10 (μ^{2} + σ^{2}) = 10 (\frac{36}{25} + \frac{14}{25}) = 20$

Q 18 :

If an unbiased die, marked with −2, −1, 0, 1, 2, 3 on its faces, is thrown five times, then the probability that the product of the outcomes is positive, is [2023]

$\frac{440}{2592}$
$\frac{881}{2592}$
$\frac{27}{288}$
$\frac{521}{2592}$

1

2

3

4

(4)

The markings on an unbiased die are −2, −1, 0, 1, 2, 3.

$P (positive number) = \frac{3}{6} = \frac{1}{2}$

$P (negative number) = \frac{2}{6} = \frac{1}{3}$

Let $A =$ Product is positive.

$A = (5 positive numbers) or (1 positive, 4 negative) or (3 positive, 2 negative)$

$P (A) = {}^{5}C_{5} {(\frac{1}{2})}^{5} + {}^{5}C_{1} {(\frac{1}{3})}^{4} {(\frac{1}{2})}^{1} + {}^{5}C_{3} {(\frac{1}{2})}^{3} \times {(\frac{1}{3})}^{2}$

$= \frac{1}{32} + \frac{5}{162} + \frac{10}{72} = \frac{521}{2592}$

Q 19 :

The probability distribution of a random variable X is given below :

X 4k $\frac{30}{7}$ k $\frac{32}{7}$ k $\frac{34}{7}$ k $\frac{36}{7}$ k $\frac{38}{7}$ k $\frac{40}{7}$ k 6k

P(X) $\frac{2}{15}$ $\frac{1}{15}$ $\frac{2}{15}$ $\frac{1}{5}$ $\frac{1}{15}$ $\frac{2}{15}$ $\frac{1}{5}$ $\frac{1}{15}$

If E(X) $= \frac{263}{15}$ then P(X < 20) is equal to : [2026]

$\frac{8}{15}$
$\frac{11}{15}$
$\frac{3}{5}$
$\frac{14}{15}$

1

2

3

4

(2)

$E (X) = \sum x_{i} P (X_{i}) = \frac{526 k}{15 \times 7} = \frac{263}{15} \Rightarrow k = \frac{7}{2}$

X	14	15	16	17	18	19	20	21
P(X)	$\frac{2}{15}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$

$P (X < 20) = \sum_{x = 14}^{19} P (X) = \frac{11}{15}$

Q 20 :

If a random variable $x$ has the probability distribution

$x$ 0 1 2 3 4 5 6 7

$P (x)$ 0 $2 k$ $k$ $3 k$ $2 k^{2}$ $2 k$ $k^{2} + k$ $7 k^{2}$

then $P (3 < x \leq 6)$ is equal to [2026]

0.64
0.34
0.33
0.22

1

2

3

4

(3)

$\sum P (x_{i}) = 1$

$\Rightarrow 9 k + 10 k^{2} = 1$

$\Rightarrow 10 k^{2} + 9 k - 1 = 0 \Rightarrow k = \frac{1}{10}$

$P (3 < x \leq 6) = 3 k + 3 k^{2}$

$= \frac{3}{10} + \frac{3}{100} = 0.33$

$= 0.33$

Topic Question Set

Q 1 :

If the mean of the following probability distribution of a random variable X :

X 0 2 4 6 8

P(X) $a$ $2 a$ $a + b$ $2 b$ $3 b$

is $\frac{46}{9},$ then the variance of the distribution is [2024]

Q 2 :

Three rotten apples are accidentally mixed with fifteen good apples. Assuming the random variable $x$ to be the number of rotten apples in a draw of two apples, the variance of $x$ is [2024]

Q 3 :

From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. If the variance of $X$ is $σ^{2},$ then $96 σ^{2}$ is equal to _____________ . [2024]

Q 6 :

A fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required and let $a = P (X = 3), b = P (X \geq 3)$ and $c = P (X \geq 6 ∣ X > 3) .$ Then $\frac{b + c}{a}$ is equal to _______ . [2024]

Q 7 :

If the probability that the random variable X takes the value x is given by $P (X = x) = k (x + 1) 3^{- x}$ , x = 0, 1, 2, 3, ..., where k is a constant, then $P (X \geq 3)$ , is equal to [2025]

Q 8 :

Let a random variable X take values 0, 1, 2, 3, with P(X = 0) = P(X = 1) = p, P(X = 2) = P(X = 3) and $E (X^{2}) = 2 E (X)$ . Then the value 8p – 1 is : [2025]

Q 9 :

Let $A = [a_{i j}]$ be a $2 \times 2$ matrix such that $a_{i j} \in {0, 1}$ for all i and j. Let the random variable X denote the possible values of the determinant of the matrix A. Then, the variance of X is : [2025]

Q 10 :

A pair of dice is thrown 5 times. For each throw, a total of 5 is considered a success. If the probability of at least 4 successes is $\frac{k}{3^{11}}$ , then $k$ is equal to [2023]

Q 11 :

If the probability that the random variable X takes values $x$ is given by $P (X = x) = k (x + 1) 3^{- x}, x = 0, 1, 2, 3, \dots$ , where $k$ is a constant, then $P (X \geq 2)$ is equal to [2023]

Q 12 :

Let a die be rolled $n$ times. Let the probability of getting odd numbers seven times be equal to the probability of getting odd numbers nine times. If the probability of getting even numbers twice is $\frac{k}{2^{15}}$ , then $k$ is equal to [2023]

Q 13 :

Two dice A and B are rolled. Let the numbers obtained on A and B be $α$ and $β$ respectively. If the variance of $α - β$ is $\frac{p}{q}$ , where $p$ and $q$ are co-prime, then the sum of the positive divisors of $p$ is equal to [2023]

Q 14 :

A coin is biased so that the head is 3 times as likely to occur as tail. This coin is tossed until a head or three tails occur. If $X$ denotes the number of tosses of the coin, then the mean of $X$ is [2023]

Q 15 :

The random variable $X$ follows binomial distribution $B (n, p)$ , for which the difference of the mean and the variance is 1. If $2 P (X = 2) = 3 P (X = 1),$ then $n^{2} P (X > 1)$ is equal to [2023]

Q 16 :

In a binomial distribution $B (n, p)$ , the sum and the product of the mean and the variance are 5 and 6 respectively, then $6 (n + p - q)$ is equal to [2023]

Q 18 :

If an unbiased die, marked with −2, −1, 0, 1, 2, 3 on its faces, is thrown five times, then the probability that the product of the outcomes is positive, is [2023]

Q 20 :

If a random variable $x$ has the probability distribution

$x$ 0 1 2 3 4 5 6 7

$P (x)$ 0 $2 k$ $k$ $3 k$ $2 k^{2}$ $2 k$ $k^{2} + k$ $7 k^{2}$

then $P (3 < x \leq 6)$ is equal to [2026]

Want Us to Email you About Special Offers & Updates?

X	4k	$\frac{30}{7}$ k	$\frac{32}{7}$ k	$\frac{34}{7}$ k	$\frac{36}{7}$ k	$\frac{38}{7}$ k	$\frac{40}{7}$ k	6k
P(X)	$\frac{2}{15}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$

$x$	0	1	2	3	4	5	6	7
$P (x)$	0	$2 k$	$k$	$3 k$	$2 k^{2}$	$2 k$	$k^{2} + k$	$7 k^{2}$

Topic Question Set

Q 1 : If the mean of the following probability distribution of a random variable X : X 0 2 4 6 8 P(X) a 2a a+b 2b 3b is 469, then the variance of the distribution is [2024]

Q 2 : Three rotten apples are accidentally mixed with fifteen good apples. Assuming the random variable x to be the number of rotten apples in a draw of two apples, the variance of x is [2024]

Q 3 : From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable X denote the number of defective items in the sample. If the variance of X is σ2, then 96σ2 is equal to _____________ . [2024]

Q 5 : Three balls are drawn at random from a bag containing 5 blue and 4 yellow balls. Let the random variables X and Y respectively denote the number of blue and yellow balls. If X¯ and Y¯ are the means of X and Y respectively, then 7X¯+4Y¯ is equal to ________ . [2024]

Q 6 : A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required and let a=P(X=3), b=P(X≥3) and c=P(X≥6∣X>3). Then b+ca is equal to _______ . [2024]

Q 7 : If the probability that the random variable X takes the value x is given by P(X=x)=k(x+1)3–x, x = 0, 1, 2, 3, ..., where k is a constant, then P(X≥3), is equal to [2025]

Q 8 : Let a random variable X take values 0, 1, 2, 3, with P(X = 0) = P(X = 1) = p, P(X = 2) = P(X = 3) and E(X2)=2E(X). Then the value 8p – 1 is : [2025]

Q 9 : Let A=[aij] be a 2×2 matrix such that aij∈{0,1} for all i and j. Let the random variable X denote the possible values of the determinant of the matrix A. Then, the variance of X is : [2025]

Q 10 : A pair of dice is thrown 5 times. For each throw, a total of 5 is considered a success. If the probability of at least 4 successes is k311, then k is equal to [2023]

Q 11 : If the probability that the random variable X takes values x is given by P(X=x)=k(x+1)3-x, x=0,1,2,3,…, where k is a constant, then P(X≥2) is equal to [2023]

Q 12 : Let a die be rolled n times. Let the probability of getting odd numbers seven times be equal to the probability of getting odd numbers nine times. If the probability of getting even numbers twice is k215, then k is equal to [2023]

Q 13 : Two dice A and B are rolled. Let the numbers obtained on A and B be α and β respectively. If the variance of α-β is pq, where p and q are co-prime, then the sum of the positive divisors of p is equal to [2023]

Q 14 : A coin is biased so that the head is 3 times as likely to occur as tail. This coin is tossed until a head or three tails occur. If X denotes the number of tosses of the coin, then the mean of X is [2023]

Q 15 : The random variable X follows binomial distribution B(n,p), for which the difference of the mean and the variance is 1. If 2P(X=2)=3P(X=1), then n2P(X>1) is equal to [2023]

Q 16 : In a binomial distribution B(n,p), the sum and the product of the mean and the variance are 5 and 6 respectively, then 6(n+p−q) is equal to [2023]

Q 17 : Three rotten apples are mixed accidently with seven good apples and four apples are drawn one by one without replacement. Let the random variable X denote the number of rotten apples. If μ and σ2 represent mean and variance of X, respectively, then 10(μ2+σ2) is equal to [2023]

Q 18 : If an unbiased die, marked with −2, −1, 0, 1, 2, 3 on its faces, is thrown five times, then the probability that the product of the outcomes is positive, is [2023]

Q 19 : The probability distribution of a random variable X is given below : X 4k 307k 327k 347k 367k 387k 407k 6k P(X) 215 115 215 15 115 215 15 115 If E(X) =26315 then P(X < 20) is equal to : [2026]

Q 20 : If a random variable x has the probability distribution x 0 1 2 3 4 5 6 7 P(x) 0 2k k 3k 2k2 2k k2+k 7k2 then P(3<x≤6) is equal to [2026]

Want Us to Email you About Special Offers & Updates?

Q 1 :

If the mean of the following probability distribution of a random variable X :

X 0 2 4 6 8

P(X) $a$ $2 a$ $a + b$ $2 b$ $3 b$

is $\frac{46}{9},$ then the variance of the distribution is [2024]

Q 2 :

Three rotten apples are accidentally mixed with fifteen good apples. Assuming the random variable $x$ to be the number of rotten apples in a draw of two apples, the variance of $x$ is [2024]

Q 3 :

From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. If the variance of $X$ is $σ^{2},$ then $96 σ^{2}$ is equal to _____________ . [2024]

Q 6 :

A fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required and let $a = P (X = 3), b = P (X \geq 3)$ and $c = P (X \geq 6 ∣ X > 3) .$ Then $\frac{b + c}{a}$ is equal to _______ . [2024]

Q 7 :

If the probability that the random variable X takes the value x is given by $P (X = x) = k (x + 1) 3^{- x}$ , x = 0, 1, 2, 3, ..., where k is a constant, then $P (X \geq 3)$ , is equal to [2025]

Q 8 :

Let a random variable X take values 0, 1, 2, 3, with P(X = 0) = P(X = 1) = p, P(X = 2) = P(X = 3) and $E (X^{2}) = 2 E (X)$ . Then the value 8p – 1 is : [2025]

Q 9 :

Let $A = [a_{i j}]$ be a $2 \times 2$ matrix such that $a_{i j} \in {0, 1}$ for all i and j. Let the random variable X denote the possible values of the determinant of the matrix A. Then, the variance of X is : [2025]

Q 10 :

A pair of dice is thrown 5 times. For each throw, a total of 5 is considered a success. If the probability of at least 4 successes is $\frac{k}{3^{11}}$ , then $k$ is equal to [2023]

Q 11 :

If the probability that the random variable X takes values $x$ is given by $P (X = x) = k (x + 1) 3^{- x}, x = 0, 1, 2, 3, \dots$ , where $k$ is a constant, then $P (X \geq 2)$ is equal to [2023]

Q 12 :

Let a die be rolled $n$ times. Let the probability of getting odd numbers seven times be equal to the probability of getting odd numbers nine times. If the probability of getting even numbers twice is $\frac{k}{2^{15}}$ , then $k$ is equal to [2023]

Q 13 :

Two dice A and B are rolled. Let the numbers obtained on A and B be $α$ and $β$ respectively. If the variance of $α - β$ is $\frac{p}{q}$ , where $p$ and $q$ are co-prime, then the sum of the positive divisors of $p$ is equal to [2023]

Q 14 :

A coin is biased so that the head is 3 times as likely to occur as tail. This coin is tossed until a head or three tails occur. If $X$ denotes the number of tosses of the coin, then the mean of $X$ is [2023]

Q 15 :

The random variable $X$ follows binomial distribution $B (n, p)$ , for which the difference of the mean and the variance is 1. If $2 P (X = 2) = 3 P (X = 1),$ then $n^{2} P (X > 1)$ is equal to [2023]

Q 16 :

In a binomial distribution $B (n, p)$ , the sum and the product of the mean and the variance are 5 and 6 respectively, then $6 (n + p - q)$ is equal to [2023]

Q 18 :

If an unbiased die, marked with −2, −1, 0, 1, 2, 3 on its faces, is thrown five times, then the probability that the product of the outcomes is positive, is [2023]

Q 20 :

If a random variable $x$ has the probability distribution

$x$ 0 1 2 3 4 5 6 7

$P (x)$ 0 $2 k$ $k$ $3 k$ $2 k^{2}$ $2 k$ $k^{2} + k$ $7 k^{2}$

then $P (3 < x \leq 6)$ is equal to [2026]