david1983 wrote:

John throws a coin until a series of three consecutive heads or three consecutive tails appears. What is the probability that the game will end after the fourth throw?

(A) 1/16

(B) 2/16

(C) 3/16

(D) 7/16

(E) 9/16

Great question! I'm happy to help.

Let's consider the routes that would lead to ending the game on the fourth toss. In order to be a win at that point, tosses #2 & #3 & #4 would all have to be the same, either H or T, and in order for the game NOT to end on the third toss, the first toss would have to be different from the other four. Thus, the only two possible sequences are

H-T-T-T ----> P = (1/2)^4 = 1/16

T-H-H-H ----> P = (1/2)^4 = 1/16

Either one of these would satisfy the condition, so we could have one or the other. OR means add in probability.

1/16 + 1/16 = 2/16 = 1/8

Answer

(B). Incidentally, the GMAT would NEVER give the unsimplified fraction 2/16 --- it would always give the simplified form 1/8.

Does all this make sense?

Mike

_________________

Mike McGarry

Magoosh Test Prep