PiyushK wrote:
A and B in turns, throw a dice. If A gets a sum of 8 before B gets a sum of 9, then A wins. But if B gets a sum on 9 before A gets a sum of 8, then B wins. Find the chances of winning of A.
A. 31/36
B. 31/76
C. 5/36
D. 45/76
E. 1/9
A. 31/36
B. 31/76
C. 5/36
D. 45/76
E. 1/9
........................
Solution:
Let, A throws the dice 1st.
For the following combinations we can get an 8.
6+2
5+3
4+4
3+5
2+6
So the probability of getting an 8 in consecutively thrown 2 dices by A= 5/36 = Probability of winning of A and losing of B (B will not get any chance here because A scored the 8 1st, thats why B is out of calculation) Like this : A B A
Next if B throws the dice 1st then this combinations will give a 9.
3+6
4+5
5+4
6+3
So the probability of getting a 9 in consecutively thrown 2 dices by B = 4/36
And probability of losing A = 1 5/36 = 31/36
Again, the probability of winning of B and losing of A = 4/36 31/36 = 1/9 31/36
So the conditional probability of winning of A = (5/36) / (1/9 31/36 + 5/36)
= 45/76 (Answer D)