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Explain Solution R.D. Sharma Class 12 Chapter 30 Probability  Exercise 30.5 Question 36 Maths Textbook Solution.

Answers (1)

Answer:\frac{5}{17}

Hint: 1+a+a^{2}+..........=\frac{1}{1-a}

Given: A and B throw a pair of dice alternately.

A wins the game if he gets a total of 7 and B wins if he gets 10

Solution: Total of 7 on dice can be obtained in following ways:

(1,6) , (6,1) , (2,5) , (5,2) , (3,4) , (4,3)

Probability\: of\: getting\: 7=\frac{6}{36}=\frac{1}{6}

Probability of not getting 7=1-\frac{1}{6}=\frac{5}{6}

Total 10 on dice : (4,6),(6,4),(5,5)

Probability of getting 10=\frac{3}{36}=\frac{1}{12}

Probability of not getting 10=1-\frac{1}{12}=\frac{11}{12}

E=getting7,F=getting10

P(\mathrm{E})=\frac{1}{6}, P(\overline{\mathrm{E}})=\frac{5}{6}, P(\mathrm{~F})=\frac{1}{12}, P(\overline{\mathrm{F}})=\frac{11}{12}

Probabilities of getting 7 in first throw=\frac{1}{6}

\begin{aligned} &\therefore P(\text { getting } 7 \text { in third throw })=P(\bar{E}) P(\bar{F}) P(E)=\frac{5}{6} \times \frac{1}{6} \times \frac{11}{12} \\\\ &P(\text { getting } 7 \text { in fifth throw })=P(\bar{E}) P(\bar{F}) P(E)=\frac{5}{6} \times \frac{11}{12} \times \frac{1}{6} \\\\ &P(\text { winning } A)=\frac{1}{6}+\left(\frac{5}{6} \times \frac{11}{12} \times \frac{1}{6}\right)+\left(\frac{5}{6} \times \frac{11}{12} \times \frac{5}{6} \times \frac{11}{12} \times \frac{1}{6}\right)+\ldots . . \end{aligned}

\begin{aligned} &\frac{\frac{1}{6}}{1-\frac{5}{6} \times \frac{11}{12}}=\frac{12}{17} \\\\ &\therefore P(\text { winning } B)=1-P(\text { winning } A) \\\\ &=1-\frac{12}{17} \\\\ &=\frac{5}{17} \end{aligned}

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