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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 24 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 24 maths textbook solution

Answers (1)

Answer: 

X 0 1 2
P\left ( X \right ) \frac{4}{9} \frac{4}{9} \frac{1}{9}


Hint: use probability formula

Given: A fair die is tossed twice if the number appearing on the top less than 3 it is successes find the probability distribution of number of successes

Solution: A fair dice is tossed twice. Every time a throwing dice is an independent event.

Note: P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right ) where A and B are independent events.

Let A denote the event of getting a number less than 3 on a single throw of dice

\therefore P\left ( A \right )=\frac{2}{6}=\frac{1}{3} { out of 6 outcomes 1 and 2 are favourable}

P(\operatorname{not} A)=P\left(A^{\prime}\right)=1-\frac{1}{3}=\frac{2}{3}

As success is considered when number appearing on dice is less than 3

Let x denotes the success

As we are throwing two dice so that we can or we may even not get success

∴x can take value 0,1 or 2

\begin{aligned} &P(x=0)=P(\text { not success }) \times \mathrm{P}(\text { not success })=P\left(A^{\prime}\right) \times P\left(A^{\prime}\right)=\frac{2}{3} \times \frac{2}{3}=\frac{4}{9} \\ &P(x=0)=P\left(A^{\prime}\right) \times P(A)+P\left(A^{\prime}\right) \times P(A) \\ &=\frac{2}{3} \times \frac{1}{3}+\frac{1}{3} \times \frac{2}{3}=\frac{4}{9} \\ &P(x=2)=P(\text { success }) \times P(\text { success })=P(A) \times P(A)=\frac{1}{3} \times \frac{1}{3}=\frac{1}{9} \end{aligned}

Now we have X and P(X)                

∴ Now we are ready to write the probability distribution of x.

The following table gives probability distribution

X 0 1 2
P\left ( X \right ) \frac{4}{9} \frac{4}{9} \frac{1}{9}
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