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

Question

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

A. 1/36
B. 1/24
C. 1/18
D. 1/12
E. 1/6

nick1816 · Answer

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

jackfr2 · Answer

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

nick1816 · Answer

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.

ScottTargetTestPrep · Answer

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

LeoN88 · Answer

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.

Regor60 · Answer

To satisfy the question stem the beginning order has to be THT.

This can proceed alternating to infinity where HH is finally achieved. So a few runs of the series looks like:

THTHH with (1/2)^5 probability 
THTHTHH with (1/2)^7 probability 
THTHTHTHH with (1/2)^9 probability

So this infinite series sums to S, the value we're looking for.

If we multiply S by 4, we get 4S = (1/2)^3 + (1/2)^5 +(1/2)^7... This looks like S but with (1/2)^3 added, so:

4S = S + 1/8 or 3S = 1/8 and S = 1/24

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards