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]

0%

(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]

0%

(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]

0%

(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]

0%

(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,

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

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

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

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

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

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

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]

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0%

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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]

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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]

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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]

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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]

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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]