Bunuel wrote:
In a diving competition, each diver has a 20% chance of a perfect dive. The first perfect dive of the competition, but no subsequent dives, will receive a perfect score. If Janet is the third diver to dive, what is her chance of receiving a perfect score? (Assume that each diver can perform only one dive per turn.)
A. 1/5
B. 1/15
C. 4/25
D. 16/125
E. 61/125
Kudos for a correct solution.
I'm not sure why they worded the question this way. As I'd understand the question, as presented, the only answer I could give is "greater than 16/125". Janet certainly has a (4/5)(4/5)(1/5) = 16/125 probability of getting a perfect score on her first dive. But as they've worded the question, it sounds like each diver will dive multiple times - the comment in brackets suggests the divers will continue diving over the course of several "turns", and the first part of the question suggests that someone "will" get a perfect score. So I'd interpret that to mean the competition will continue until someone gets a perfect score, and that means Janet will potentially have many chances at a perfect dive. We'd need to know how many divers there are in total to then give an exact answer to the question. If there were three divers, and they kept diving until someone got a perfect score, the answer would be
(4/5)(4/5)(1/5) + (4/5)(4/5)(4/5)(4/5)(4/5)(1/5) + ...
= (4/5)^2 (1/5) + (4/5)^5 (1/5) + (4/5)^8 (1/5) + ...
= (16/125) (1 + (4/5)^3 + (4/5)^6 + ...)
The question is now beyond the scope of the GMAT -- that's an infinite geometric series in brackets, and you don't need to know any formulas for those. But using the geometric series formula, the sum in brackets adds to 1/(1 - 16/125) = 125/109, so the answer is 16/109 in this case. With more divers, the answer would be smaller, but still greater than 16/125.