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## Kate and Danny each have $10. Together, they flip a fair coi ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult Kate and Danny each have$10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives Danny $1. Every time the coin lands on tails, Danny gives Kate$1. After the five coin flips, what is the probability that Kate has more than $10 but less than$15?

(A) 5/16
(B) 1/2
(C) 12/30
(D) 15/32
(E) 3/8

OA D

Source: Manhattan Prep

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BTGmoderatorDC wrote:
Kate and Danny each have $10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives Danny$1. Every time the coin lands on tails, Danny gives Kate $1. After the five coin flips, what is the probability that Kate has more than$10 but less than $15? (A) 5/16 (B) 1/2 (C) 12/30 (D) 15/32 (E) 3/8 OA D Source: Manhattan Prep In order for Kate to have more than 10 dollars but less than 15 dollars, either of the following two outcomes must have occurred: T-T-T-T-H, so Kate would have 13 dollars T-T-T-H-H, so Kate would have 11 dollars Letâ€™s calculate the probability of each outcome: P(T-T-T-T-H) = (1/2)^5 = 1/32 Since T-T-T-T-H can be arranged in 5!/4! = 5 ways, the probability is 5/32. Next: P(T-T-T-H-H) = (1/2)^5 = 1/32 Since T-T-T-H-H can be arranged in 5!/(3! x 2!) = 10 ways, the probability is 10/32. So the overall probability that Kate has more than$10 but less than $15 is 15/32. Answer: D _________________ Scott Woodbury-Stewart Founder and CEO scott@targettestprep.com See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews ### Top Member Master | Next Rank: 500 Posts Joined 15 Oct 2009 Posted: 330 messages Upvotes: 27 Top Reply BTGmoderatorDC wrote: Kate and Danny each have$10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives Danny $1. Every time the coin lands on tails, Danny gives Kate$1. After the five coin flips, what is the probability that Kate has more than $10 but less than$15?

(A) 5/16
(B) 1/2
(C) 12/30
(D) 15/32
(E) 3/8

OA D

Source: Manhattan Prep
Since the payoffs are equal, one can quickly see that Kate has a 50% chance of having more than what she started with, $10. But that includes the excluded case of having$15. To win an additional $5, she'd have to win every toss, which has a probability of (1/2)^5 or 1/32. So the probability is 1/2 = 16/32 - 1/32 = D, 15/32 ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15362 messages Followed by: 1866 members Upvotes: 13060 GMAT Score: 790 BTGmoderatorDC wrote: Kate and Danny each have$10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives Danny $1. Every time the coin lands on tails, Danny gives Kate$1. After the five coin flips, what is the probability that Kate has more than $10 but less than$15?

(A) 5/16
(B) 1/2
(C) 12/30
(D) 15/32
(E) 3/8
For every tails, Kate gets $1; for every heads, she loses$1.

0 tails, 5 heads: Kate has 10+0-5 = $5 1 tails, 4 heads: Kate has 10+1-4 =$7
2 tails, 3 heads: Kate has 10+2-3 = $9 3 tails, 2 heads: Kate has 10+3-2 =$11
4 tails, 1 heads: Kate has 10+4-1 = $13 5 tails, 0 heads: Kate has 10+5-0 =$15

Only 3 tails and 4 tails will give Kate between $10 and$15.

Question stem, rephrased:
What is the probability that coin flipped 5 times will yield exactly 3 tails or exactly 4 tails?

When a coin is flipped 5 times, we have an equal chance of getting tails 3, 4 or 5 times as we do of getting tails 0, 1 or 2 times.
Thus, P(3 tails) + P(4 tails) + P(5 tails) = 1/2.
P(5 tails) = 1/32.
So P(3 tails) + P(4 tails) = 1/2 - 1/32 = 15/32.

The correct answer is B.

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