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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 25 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 25 maths textbook solution

Answers (1)

Answer:

X 0 1 2
P\left ( X \right ) \frac{10}{21} \frac{10}{21} \frac{1}{21}


Given: An urn contains 5 red and 2 black balls, two balls are randomly selected. let X represent the number of black balls. What are the possible values of x? is x a random variable?

Solution: the key point to solve the problem: A variable X is said to be a random variable if the sum of probabilities associated with each value of X gets equal to 1

i.e. \varepsilon \left ( p_{1} \right )=1 where p_{1} is probability associated with x_{1}

x represents the number of black balls drawn

∴ x can take value 0,1 or 2 as both balls drawn can be black which corresponds to x = 2, either of 2 balls can be black which corresponds to x = 1 and if neither of balls drawn is black it corresponds to  x = 0 

∴ there are the of 7 balls.

\mathrm{n}(\mathrm{s})=$ total possible ways of selecting 2 balls $=7 c_{2}

P(x=0)=P($ selecting no blank balls $)

\frac{5 c_{2}}{7 c_{2}}=\frac{5 \times 4}{7 \times 6}=\frac{10}{21}

P(x=1)=P($ selecting 1 black ball and 1 red $)

=\frac{5 c_{1} \times 2 c_{1}}{7 c_{2}}=\frac{5 \times 2 \times 2}{7 \times 6}=\frac{10}{21}

P(x=2)=P($ selecting all black balls $)

=\frac{2 c_{2}}{7 c_{2}}=\frac{2}{7 \times 6}=\frac{1}{21}

X 0 1 2
P\left ( X \right ) \frac{10}{21} \frac{10}{21} \frac{1}{21}

Clearly, \varepsilon\left(P_{1}\right)=\frac{10}{21}+\frac{10}{21}+\frac{1}{21}=1

∴ x is a random variable, and the above table represents its probability distribution.

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