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  • Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31 point 1 question 12 maths textbook solution

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

Answers (1)

Answer:

X 2 3 4 5 6 7 8 9 10 11 12
P\left ( X \right ) \frac{1}{36} \frac{1}{18} \frac{1}{12} \frac{1}{9} \frac{5}{36} \frac{1}{6} \frac{5}{36} \frac{1}{9} \frac{1}{12} \frac{1}{18} \frac{1}{36}


Hint: Use probability distribution formula

Given: 2 dice are thrown together and the no. was noted x, denotes the sum of the 2 nos. Assuming that all the 36 outcomes are equally likely, we have to find the probability distribution of x

Solution: When two fair dice are thrown there are total 36 possible outcomes.

\therefore \lambda denotes the sum of 2 numbers appearing on dice.

\therefore x can take values 2,3,4,5,6,7,8,9,10,11,12   As appearance of a number on a fair die is equally likely

\begin{aligned} &\text { i. e. } P(\text { appearing of } 1)=P(\text { appearing of } 2)=P(\text { appearing of } 3)=P(\text { appearing of } 4) \\ &=P(\text { appearing of } 5)=P(\text { appearing of } 6)=\frac{1}{6} \end{aligned}

And also the appearance of numbers on two different dice is an independent event: so two find conditions like P(1 in the first dice and 2 in the second dice) can be given using multiplication rule of probability.

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

P(x=2)=\frac{1}{36}\{\therefore(1,1)$ is the only combination resulting sum $=2\}

P(x=3)=\frac{2}{36}=\frac{1}{18}\quad\{\therefore(1,2)$ and $(2,1)$ are the combination resulting in sum $=3\}

P(x=4)=\frac{3}{36}=\frac{1}{12} \quad\{\therefore(1,3)(3,1)$ and $(2,2)$ are the combination resulting in sum $=4\}

P(x=5)=\frac{4}{36}=\frac{1}{9}\quad\{\therefore(3,2)(2,3)(1,4)$ and $(4,1)$ are the combination resulting in sum $=5\}

P(x=6)=\frac{5}{36}\{\therefore(1,5)(5,1)(2,4)(4,2)(3,3)$ are the combination resulting in sum $=6\}

P(x=7)=\frac{6}{36}=\frac{1}{6}

\{\therefore(1,6)(6,1)(2,5)(5,2)(3,4)(4,3) are the combination resulting in sum=7\}

P(x=9)=\frac{4}{36}=\frac{1}{9} \quad\{\therefore(3,6)(6,3)(5,4)(4,5)$ are the combination resulting in sum $=9\}

P(x=10)=\frac{3}{36}=\frac{1}{12} \quad\{\therefore(6,4)(4,6)$ and $(5,5)$ are the combination resulting in sum $=10\}

P(x=11)=\frac{2}{36}=\frac{1}{18} \quad\{\therefore(5,6)$ and $(6,5)$ are the combination resulting in sum $=11\}

P(x=12)=\frac{1}{36} \quad\{\therefore(6,6)$ is the only combination resulting in sum $=2\}

Now we have X and P(X)

\therefore Required probablity distribution is

X 2 3 4 5 6 7 8 9 10 11 12
P\left ( X \right ) \frac{1}{36} \frac{1}{18} \frac{1}{12} \frac{1}{9} \frac{5}{36} \frac{1}{6} \frac{5}{36} \frac{1}{9} \frac{1}{12} \frac{1}{18} \frac{1}{36}
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