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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 30 Probability Exercise 30.5 Question 26 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 30 Probability  Exercise 30.5 Question 26 Maths Textbook Solution.

Answers (1)

Answer:

P\left ( A\: winning \right )=\frac{4}{7}

P\left ( B\: winning \right )=\frac{2}{7}

P\left ( C\: winning \right )=\frac{1}{7}

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

Given: A, B, C are in order. The one to throw a head wins.

Solution: P(A winning) = P(head in first toss)+P(head in 4th toss)+….

\begin{aligned} &=\frac{1}{2}+\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right)+\ldots \ldots \\ &=\frac{1}{2}\left[1+\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{6}+\ldots\right] \\ &=\frac{1}{2}\left[\frac{1}{1-\left(\frac{1}{2}\right)^{3}}\right] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad\left[1+a+a^{2}+a^{3}+\ldots=\frac{1}{1-a}\right] \\ \end{aligned}

\begin{aligned} &=\frac{1}{2}\left[\frac{1}{1-\frac{1}{8}}\right] \\ &=\frac{1}{2} \times \frac{8}{7} \\ &=\frac{4}{7} \end{aligned}

P(B winning)= P(head in second toss)+P(head in 5th toss)+….

\begin{aligned} &=\left(\frac{1}{2} \times \frac{1}{2}\right)+\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right)+\ldots \ldots . \\ &=\frac{1}{4}\left[1+\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{6}+\ldots\right] \\ &=\frac{1}{4}\left[\frac{1}{1-\left(\frac{1}{2}\right)^{3}}\right] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad\left[1+a+a^{2}+a^{3}+\ldots=\frac{1}{1-a}\right] \\ \end{aligned}

\begin{aligned} &=\frac{1}{4}\left[\frac{1}{1-\frac{1}{8}}\right] \\\\ &=\frac{1}{4} \times \frac{8}{7} \\\\ &=\frac{2}{7} \end{aligned}

P (C winning)= P(head in third toss)+P(head in 6th toss)+….

\begin{aligned} &=\left(\frac{1}{2} \times \frac{1}{2}\times \frac{1}{2}\right)+\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\times \frac{1}{2}\right)+\ldots \ldots . \\ &=\frac{1}{8}\left[1+\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{6}+\ldots\right] \\ &=\frac{1}{8}\left[\frac{1}{1-\left(\frac{1}{2}\right)^{3}}\right] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad\left[1+a+a^{2}+a^{3}+\ldots=\frac{1}{1-a}\right] \\ \end{aligned}

\begin{aligned} &=\frac{1}{8}\left[\frac{1}{1-\frac{1}{8}}\right] \\\\ &=\frac{1}{8} \times \frac{8}{7} \\\\ &=\frac{1}{7} \end{aligned}

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