Bunuel wrote:
Debra flips a fair coin repeatedly, keeping track of how many heads and how many tails she has seen in total, until she gets either two heads in a row or two tails in a row, at which point she stops flipping. What is the probability that she gets two heads in a row but she sees a second tail before she sees a second head?
A. 1/36
B. 1/24
C. 1/18
D. 1/12
E. 1/6
The ways she gets two heads in a row is:
HH, THH, HTHH, THTHH, HTHTHH, THTHTHH, HTHTHTHH, THTHTHTHH, ...
The probability of getting each of the above cases is 1/4, 1/8, 1/16, 1/32, 1/64, 1/128, 1/256, 1/512, ....
Of the cases listed above, only the bold ones have a second tail before a second head; and the probability of getting them is also in bold, i.e., the numbers 1/32, 1/128, 1/512, …. This is also an infinite geometric series. Recall that the sum for such a series is a/(1 - r) where a is the first term and r is the common ratio. Here, a = 1/32 and r = 1/4; therefore, the sum of these probabilities (the numbers in bold) is (1/32)/(1 - 1/4) = (1/32)/(3/4) = 1/24.
Answer: B
_________________