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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 30 Probability Exercise 30.5 Question 28 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 30 Probability  Exercise 30.5 Question 28 Maths Textbook Solution.

Answers (1)

Answer:Proved

Hint: You must know the rules of finding probability functions.

Given: A and B take turns in throwing two dice,

The first throw 10 being awarded the prize.

Solution: There are only three possible cases, where the sum of number obtained after throwing

2 dice is 10, i.e.[(4,6) , (5,5) , (6,4)]

P( sum \: of \: number\: is 10)=\frac{3}{36}=\frac{1}{12}

P( sum \: of \: number \: is \: not\: 10)=1-\frac{1}{12}=\frac{11}{12}

P(\mathrm{~A} \: winning )=P(10 \: in \: first \: throw)+P(10 \: in\: third \: throw )+\ldots

\begin{aligned} &=\frac{1}{12}+\left(\frac{11}{12} \times \frac{11}{12} \times \frac{1}{12}\right)+\ldots \ldots \\ &=\frac{1}{12}\left[1+\left(\frac{11}{12}\right)^{2}+\left(\frac{11}{12}\right)^{4}+\ldots\right] \\ &=\frac{1}{12}\left[\frac{1}{1-\frac{121}{144}}\right] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad\left[1+a+a^{2}+a^{3}+\ldots=\frac{1}{1-a}\right] \\ \end{aligned}

\begin{aligned} &=\frac{1}{12} \times \frac{144}{23} \\\\ &=\frac{12}{23} \end{aligned}

P(\mathrm{~B} \text { winning })=1-P(\mathrm{~A} \text { winning })=\frac{11}{23}

Now,

\frac{P(\mathrm{~A} \text { winning })}{P(\mathrm{~B} \text { winning })}=\frac{\frac{12}{23}}{\frac{11}{23}}=\frac{12}{11}

Posted by

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