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Need Solution for R.D.Sharma Maths Class 12 Chapter 30 Probability  Exercise 30.5 Question 25 Maths Textbook Solution.

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Answer:\frac{9}{8}

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

Given: A and B take turns in throwing two dice, the first to throw 9 being awarded the prize

Solution: Total number of events = 36

P(getting 9) =\frac{4}{36}=\frac{1}{9}

P(A winning) = P(getting 9 in first throw) + P(getting 9 in third row)+….

\begin{aligned} &=\frac{1}{9}+\left(1-\frac{1}{9}\right)\left(1-\frac{1}{9}\right) \times \frac{1}{9}+\ldots . . \\\\ &=\frac{1}{9}\left[1+\frac{64}{81}+\left(\frac{64}{81}\right)^{2}+\ldots\right] \\\\ &=\frac{1}{9}\left[\frac{1}{1-\frac{64}{81}}\right] \quad\left[1+a+a^{2}+a^{3}+\ldots=\frac{1}{1-a}\right] \\\\ &=\frac{1}{9}\left[\frac{81}{17}\right] \\\\ &=\frac{9}{17} \end{aligned}

P(B winning)=P(getting 9 in second throw)+P(getting 9 in fourth throw)+…

\begin{aligned} &=\left(1-\frac{1}{9}\right) \frac{1}{9}+\left(1-\frac{1}{9}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{9}\right) \times \frac{1}{9}+\ldots . . \\\\ &=\frac{8}{81}\left[1+\frac{64}{81}+\left(\frac{64}{81}\right)^{2}+\ldots .\right] \\\\ &=\frac{8}{81}\left[\frac{1}{1-\frac{64}{81}}\right] \quad\left[1+a+a^{2}+a^{3}+\ldots=\frac{1}{1-a}\right] \\\\ &=\frac{8}{81}\left[\frac{81}{17}\right] \\\\ &=\frac{8}{17} \end{aligned}

Winning ratio of A to B\frac{\frac{9}{17}}{\frac{8}{17}}=\frac{9}{17}\times \frac{8}{17}=\frac{9}{8}

 

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