Question

(1.5)
(b) Two players A and B toss a die alternately. He who first throws a "six" wins the
game. If A begins, what is the probability that he wins? What is the probability of B
winning the game?​

Answer 1

Answer:

11,5

Step-by-step explanation:

Let S denote the success (getting a ‘6’) and F denote the failure (not getting a ‘6’) .

Thus, P(S)=

6

1

=p, P(F)=

6

5

=q

P(A wins in first throw)=P(S)=p

P(A wins in third throw)=P(FFS)=qqp

P(A wins in fifth throw)=P(FFFFS)=qqqqp

So, P(A wins)=p+qqp+qqqqp+…

=p(1+q

2

+q

4

+…)

=

1−q

2

p

=

1−

36

25

6

1

=

11

6

P(B wins)=1–P(A wins)

P(B wins) =1−

11

6

=

11

5

So, P(A wins)=

11

6

, P(B wins)=

11

5