Solution Let S denote the success (getting a ‘6’) and F denote the failure (not getting
a ‘6’) Thus,
P(S) = 1
5
6, P(F)
6
=
P(A wins in the first throw) = P(S) = 1
6
A gets the third throw, when the first throw by A and second throw by B result into
failures Therefore,
P(A wins in the 3rd throw) = P(FFS) = 
5
5
1
P(F)P(F)P(S)= 6
6
6
 
 
=
52
1
6
6
⎛
⎞ ×
⎜
⎟
⎝
⎠
P(A wins in the 5th throw) = P (FFFFS) 
54
1
6
6
 
 
 
 
  
 
 
 
 
 
 
 and so on Hence,
P(A wins) =
2
4
1
5
1
5
1