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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 30 Probability Exercise 30.5 Question 27 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 30 Probability  Exercise 30.5 Question 27 Maths Textbook Solution.

Answers (1)

Answer:

P\left ( A\: winning \right )=\frac{36}{91}

P\left ( B\: winning \right )=\frac{30}{91}

P\left ( C\: winning \right )=\frac{25}{91}

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

Given:  Three persons A,B,C throw a die in succession till one gets a six and wins the game.

Solution:

\begin{aligned} &P(\operatorname{six})=\frac{1}{6} \\ &P(\text { no } \operatorname{six})=\frac{5}{6} \\ &P(\text { A winning })=P(6 \text { in first throw })+P(6 \text { in fourth throw })+\ldots \\ \end{aligned}

\begin{aligned} &=\frac{1}{6}+\left(\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\right)+\ldots \ldots . \\ &=\frac{1}{6}\left[1+\left(\frac{5}{6}\right)^{3}+\left(\frac{5}{6}\right)^{6}+\ldots\right] \\ \end{aligned}

\begin{aligned} &=\frac{1}{6}\left[\frac{1}{1-\frac{125}{216}}\right] \quad\left[1+a+a^{2}+a^{3}+\ldots=\frac{1}{1-a}\right] \\\\ &=\frac{1}{6} \times \frac{216}{91} \\\\ &=\frac{36}{91} \end{aligned}

\begin{aligned} &P(\mathrm{~B} \text { winning })=P(6 \text { in second throw })+P(6 \text { in fifth throw })+\ldots \\ \end{aligned}

\begin{aligned} &=\frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\right)+\ldots . . . \\\\ &=\frac{5}{36}\left[1+\left(\frac{5}{6}\right)^{3}+\left(\frac{5}{6}\right)^{6}+\ldots\right] \\ \end{aligned}

\begin{aligned} &=\frac{5}{36}\left[\frac{1}{1-\frac{125}{216}}\right] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad\left[1+a+a^{2}+a^{3}+\ldots=\frac{1}{1-a}\right] \\\\ &=\frac{5}{36} \times \frac{216}{91} \\\\ &=\frac{30}{91} \end{aligned}

 

\begin{aligned} &P(\mathrm{~C} \text { winning })=P(6 \text { in third throw })+P(6 \text { in sixth throw })+\ldots \\ \end{aligned}

\begin{aligned} &=\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\right)+\ldots \ldots \\\\ &=\frac{25}{216}\left[1+\left(\frac{5}{6}\right)^{3}+\left(\frac{5}{6}\right)^{6}+\ldots\right] \\ \end{aligned}

\begin{aligned} &=\frac{25}{216}\left[\frac{1}{1-\frac{125}{216}}\right] \quad\left[1+a+a^{2}+a^{3}+\ldots=\frac{1}{1-a}\right] \\\\ &=\frac{25}{216} \times \frac{216}{91} \\\\ &=\frac{25}{91} \end{aligned}

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