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Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 10 maths textbook solution

Answers (1)

Answer: Required probability distribution is

$X$	$0$	$1$	$2$	$3$
$P\left ( X \right )$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{3}{10}$	$\frac{1}{30}$

Hint: Let x denotes number of red balls out of 3 drawn. Then $x=0,1,2,3$

Given: A bag has 4 red and 6 black balls, three balls are drawn

Solution:

$P( no red balls )=P(x=0)=\frac{6 C 3}{10 c_{3}}=\frac{6 \times 5 \times 4}{3 \times 2}=\frac{3 \times 2}{10 \times 9 \times 8}=\frac{1}{6}$

$P( one red balls )=P(x=1)=\frac{4 C 1 \times 6 C 2}{10 C 3}=\frac{4 \times 6 \times 5}{2}=\frac{3 \times 2}{10 \times 9 \times 8}$ $=\frac{1}{2}$

$P( two red balls )=P(x=2)=\frac{4 C 2 \times 6 C 1}{10 C 3}=\frac{4 \times 3 \times 6}{2}=\frac{3 \times 2}{10 \times 9 \times 8}$ $=\frac{3}{10}$

$P( all three red )=P(x=3)=\frac{4 C 3}{10 C 3}=\frac{4 \times 3 \times 2}{10 \times 9 \times 8}=\frac{1}{30}$

The required probability distribution is

$X$	$0$	$1$	$2$	$3$
$P\left ( X \right )$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{3}{10}$	$\frac{1}{30}$

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