A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8, and B wins if he throws a sum of 8 before A throws a sum of 5. The probability, that A wins if A makes the first throw, is
- 819
- 919
- 817
- 917
Answer
(2) 919
Solution
p= 436= 19 (sum 5), q= 536 (sum 8); A wins on throw 1, 3, 5, … if both fail before:
P(A)=p+(1-p)(1-q) p+ [(1-p)(1-q) ]2p+
Common ratio r= 89 3136= 6281
P(A)= 1/91-62/81= 19 8119= 919
Answer: (2)