Aand b alternate rolling a pair of dice, stopping either when a rolls the sum 9 or when b rolls the sum 6. assuming that a rolls first, find the probability that the final roll is made by a.

There are 6*6 total combinations = 36 combinations.

For A: the probability that one rolls a 9 is 4/36

(9  = 3&6, 6&3, 4&5, 5&4) 

For B: the probability that one rolls a 6 is 5/36 

(6  = 3&3, 1&5, 5&1, 2&4, 4&2) 

So find the probability that final roll is made by A:
P(A wins) = P(A wins in first roll) + P(A wins in 3rd roll) +  
P(A wins) = 4/36 + (32/36)*(31/36)*(4/36) + (32/36)^2 *(31/36)^2 *(4/36) + ... 

Therefore we see that the common ratio is (32*31/36^2) 

Sum is = (4/36) / [1 - (32*31/36^2)] 
= 144 / 304 
= 9 / 19     (ANSWER)

true? is that what you're asking or do you mean something else?

i would say b) is your answer

sorry if im wrong

step-by-step explanation: