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Bunuel
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Sum of infinite GP= a/1-r
a is the first term
and r is the common ratio

a={1/2}^5
r= {1/2}^7/{1/2}^5= {1/2}^2
Put those values in above mentioned formula.

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= {1/2}^5/1-{1/2}^2
=1/24


can you elaborate on why are we subtracting {1/2}^2 ?
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Bunuel
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
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Bunuel
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


Two concepts are checked here:

1. For multiple probability cases we must keep adding
2. Sum of infinite GP: a/(1-r)

End of the flip will have HH for sure.
Only way she "she sees a second tail before she sees a second head"
is THT.....HH

Thus THTHH + THTHTHH + THTHTHTHH...
Now find probability.
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