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  • Explain Solution R.D. Sharma Class 12 Chapter 30 Probability Exercise 30.5 Question 35 Maths Textbook Solution.

Explain Solution R.D. Sharma Class 12 Chapter 30 Probability  Exercise 30.5 Question 35 Maths Textbook Solution.

Answers (1)

Answer: P\left ( Team\: A \right )=\frac{6}{11}

  P\left ( Team\: B \right )=\frac{5}{11}

The decision was fair as the two probabilities are almost equal.

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

Given: In a hockey match, both teams scored same number of goals upto end of the game.

Solution:

P\left ( a\: six \right )=\frac{1}{6}

P\left ( not\: a\: six \right )=\frac{5}{6}

\begin{aligned} &P(\mathrm{~A} \text { wins })=P(6 \text { in first throw })+P(6 \text { in third throw })+\ldots \\ \end{aligned}

\begin{aligned} &=\frac{1}{6}+\left(\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\right)+\ldots . . \\ &=\frac{1}{6}\left[1+\left(\frac{5}{6}\right)^{2}+\left(\frac{5}{6}\right)^{4}+\ldots\right] \\ \end{aligned}

\begin{aligned} &=\frac{1}{6}\left[\frac{1}{1-\frac{25}{36}}\right] \quad\left[1+a+a^{2}+a^{3}+\ldots=\frac{1}{1-a}\right] \\ &=\frac{1}{6} \times \frac{36}{11} \\ &=\frac{6}{11} \end{aligned}

\begin{aligned} &P(\mathrm{~B} \text { wins })=P(6 \text { in second throw })+P(6 \text { in fourth 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{1}{6}\right)+\ldots . . \\\\ &=\frac{5}{36}\left[1+\left(\frac{5}{6}\right)^{2}+\left(\frac{5}{6}\right)^{4}+\ldots .\right] \\ \end{aligned}

\begin{aligned} &=\frac{5}{36}\left[\frac{1}{1-\frac{25}{36}}\right] \quad\left[1+a+a^{2}+a^{3}+\ldots=\frac{1}{1-a}\right] \\\\ &=\frac{5}{36} \times \frac{36}{11} \\\\ &=\frac{5}{11} \end{aligned}

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