A and B throw a pair of dice 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.

Total possible outcomes for A winning the game = 6
Sample space = 36 outcomes.
P( A winning) = 1/6
Hence, A' =5/6
Total possible outcomes for B winning the game = 3
P(B winning the game) = 1/12
Hence, 11/12
Since, A starts the game therefore, the probability that B wins the game is
(5/6 x 1/12) + (5/6  11/12 x 5/6 x 1/12) + ... and so on till infinite terms.
Hence, a= 5/72 and r = 55/72
Therefore, Probability that B wins the game = 5/72 / 1-55/72 = 5/17
 

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A AND B THROW ALTERNATIVELY A PAIR OF DICE.A WINS IF HE THROWS A SUM OF 6 BEFORE B THROWS A SUM OF 7 AND B WINS IF HE THROWS A SUM OF 7 BEFORE A THROWS A SUM OF 6.FIND THE PROBABILITY OF WINNIING OF A,IF A BEGINS.
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No. of possible outcomes = 36

No. of outcomes that will result in total of 7 = 6

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

No. of outcomes that will result in total of 10= 3

[ (4,6),(5,5),(6,4)]

Probability of A getting sum of 7= P(A) = 6/36 =1/6

& Probability of B getting sum of 10 = P(B) = 3/36 = 1/12

Now, if A starts the game,

Probability of B winning =

B winning in 1st chance + B winning in 2nd chance + B winning in 3rd chance +………

=P(A’)P(B) + P(A’)P(B’)P(A’)P(B) + P(A’)P(B’)P(A’)P(B’)P(A’)P(B)+…………..

=5/6 * 1/12 + 5/6 * 11/12 * 5/6 * 1/12 + 5/6 * 11/12 * 5/6 *11/12 *5/6 *1/12 + ……………

=5/6*1/12[ 1 + 5/6*11/12 + 5/6^2 * 11/12^2 +…….]

Now , the term in the bracket is an infinite geometric series with a = 1 & r =5/6*11/12 i.e.55/72

Now the term in the bracket can be distilled to a/(1-r)

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Sir good morning
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Simplify

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5/17
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Gujjar
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Total possible outcomes for A winning the game = 6
Sample space = 36 outcomes.
P( A winning) = 1/6
Hence, A' =5/6
Total possible outcomes for B winning the game = 3
P(B winning the game) = 1/12
Hence, 11/12
Since, A starts the game therefore, the probability that B wins the game is
(5/6 x?1/12) + (5/6 ?11/12 x 5/6 x 1/12) + ... and so on till infinite terms.
Hence, a= 5/72 and r = 55/72
Therefore, Probability that B wins the game = 5/72 / 1-55/72 =?5/17
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