# A and B throw a die alternatively till one of them gets a number greater than four and wins the game. If A starts the game, what is the probability of B winning? - Mathematics

A and B throw a die alternatively till one of them gets a number greater than four and wins the game. If A starts the game, what is the probability of B winning?

#### Solution

Let S denote the success, i.e. getting a number greater than four and F denote the failure, i.e. getting a number less than four.

∴ P(S)=2/6=1/3,    P(F)=1−1/3=2/3

Now, B gets the second throw, if A fails in the first throw.

P(B wins in the second throw) = P(FS)  = P(F)P(S) = 2/3xx1/3

Similarly, P(B wins in the fourth throw) = P(FFFS)  = P(F)P(F)P(F)P(S) =  (2/3)^3xx1/3

P(B wins in the sixth throw) = P(FFFFFS)  = P(F)P(F)P(F)P(F)P(F)P(S) = (2/3)^5xx1/3 and so on.

Hence,

P(B wins) = 2/3xx1/3 + (2/3)3xx1/3 + (23)5xx1/3 + ...

=2/3xx1/3xx[1+(2/3)2+(2/3)4+......

= 2/3xx1/3xx(1/(1−4/9))   [a+ar+ar2+... =a/(1r)]

= 2/5

Thus, the probability that B wins is 2/5.

Concept: Probability Examples and Solutions
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