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

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

Answers (1)

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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