Debra flips a fair coin repeatedly, keeping track of how many heads an

Debra sees a second tail before she sees a second head>>>> She must get Tail at the first turn
And she must get two heads in a row after she she get 2 tails
The minimum number of coin flips required=5 {THTHH} and the probability={1/2}^5
The probability of getting two Heads in a row at the 6th and 7th turn= {1/2}^7
The probability of getting 2 heads in a row at 8Th and 9th turn= {1/2}^9 ..... and this pattern will go on to infinite turns

Overall probability that she gets two heads in a row but she sees a second tail before she sees a second head= {1/2}^5+{1/2}^7+{1/2}^9......infinite turns

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

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

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.

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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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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