Question

A

and

B

play a game by alternately throwing a pair of dice. One who throws

8

wins the game. If

A

starts the game, find their chances of winning.

Accepted Answer

Total number of possible outcomes in throwing a pair of dice $B$

and $8$ can be obtained in following ways $(6, 2), (5, 3), (4, 4), (3, 5), (2, 6)$

Let $P (S) =$ probability of success $=$ probability of getting $8$ in throwing a pair of dice

$P (S) = \frac{5}{36}$

and $P (F) =$ probability of failure $=$ probability of not getting $8$

$P (F) = 1 - \frac{5}{36} = \frac{31}{36}$

Probability of $A$ winning $=$ Probability of $A$ winning in first attempt or $A$ losing first and $B$ losing again and $A$ winning (this process repeats till infinity)

So, $P (A wins) = P (S) + P (F) P (F) P (S) + P (F) P (F) P (F) P (F) P (S) + . .$

$= \frac{5}{36} + \frac{31}{36} \cdot \frac{31}{36} \cdot \frac{5}{36} + \frac{31}{36} \cdot \frac{31}{36} \cdot \frac{31}{36} \cdot \frac{31}{36} \cdot \frac{5}{36} + . . .$

$= \frac{\frac{5}{36}}{1 - {(\frac{31}{36})}^{2}}$ [Since, series is in G.P with first term, $a = \frac{5}{36}$ , common ratio $r = {(\frac{31}{36})}^{2}$ , then sum of series $= \frac{a}{1 - r}$ ]

$= \frac{5}{36} \times \frac{36^{2}}{(36^{2} - 31^{2})}$

$= \frac{5}{5 \times 36} \times 36$

So, $P (A wins) = \frac{36}{67}$

Hence, $P (B wins) = 1 - P (A wins) = 1 - \frac{36}{67} = \frac{31}{67}$ .

Odisha Board Solutions for Chapter: Probability, Exercise 2: EXERCISES 6(b)

Odisha Board Mathematics Solutions for Exercise - Odisha Board Solutions for Chapter: Probability, Exercise 2: EXERCISES 6(b)

Questions from Odisha Board Solutions for Chapter: Probability, Exercise 2: EXERCISES 6(b) with Hints & Solutions

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