# A and B throw a pair of die alternately. A wins the game if he gets a total of 7 and B wins the game if he gets a total of 10. If A starts the game, then find the probability that B wins.

Given that A and B throw a pair of die alternatively.

It is told that A wins if he gets sum of 7 and B wins if he gets sum of 10.

Let us find the probability of getting sum 7 and 10.

The possibilities of getting 7 on throwing a pair of die:

{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} = 6 cases

The possibilities of getting 10 on throwing a pair of die:

{(4,6),(5,5),(6,4)} = 3 cases

P(S7) = P(getting sum 7)

P(SN7) = P(not getting sum 7)

P(SN7) = 1-P(S7)

P(S10) = P(getting sum 10)

P(SN10) = P(not getting sum 10)

P(SN10) = 1-P(S10)

It is told that A starts the game.

We need to find the probability that B wins the games.

B wins the game only when A losses in1st,3rd,5th,…… throws.

This can be shown as follows:

P(BWINS) = P(B wins the game)

P(BWins) = P(SN7S10) + P(SN7SN10SN7S10) + P(SN7SN10SN7SN10SN7S10) + …………………

Since throwing a pair of die by each person is an independent event, the probabilities multiply each other.

P(Bwins) = (P(SN7)P(S10)) + (P(SN7)P(SN10)P(SN7)P(S10)) + ………………

The terms in the bracket resembles the infinite geometric series sequence:

We know that the sum of a Infinite geometric series with first term ‘a’ and common ratio ‘r’ is

The required probability is .

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