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5/16
15/32
Â½
21/32
11/16

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rtaha2412 wrote:
Kate and David each have $10. Together they flip a coin 5 times. Every time the coin lands on heads, Kate gives David$1. Every time the coin lands on tails, David gives Kate $1. After the coin is flipped 5 times, what is the probability that Kate has more than$10 but less than $15? 5/16 15/32 Â½ 21/32 11/16 Kate will have more than 10$ and less than 15$, if she wins 3 or 4 times. So, probability of getting 3 tails or 4 tails = 5C3/2^5 + 5C4/2^5 = 10/32 + 5/32 = 15/32. B _________________ Thanks Anshu (Every mistake is a lesson learned ) ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15387 messages Followed by: 1872 members Upvotes: 13060 GMAT Score: 790 rtaha2412 wrote: Kate and David each have$10. Together they flip a coin 5 times. Every time the coin lands on heads, Kate gives David $1. Every time the coin lands on tails, David gives Kate$1. After the coin is flipped 5 times, what is the probability that Kate has more than $10 but less than$15?
5/16
15/32
Â½
21/32
11/16
As noted above, Kate will have between $10 and$15 if the coin lands on tails 3 or 4 times.

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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GMATGuruNY wrote:
rtaha2412 wrote:
Kate and David each have $10. Together they flip a coin 5 times. Every time the coin lands on heads, Kate gives David$1. Every time the coin lands on tails, David gives Kate $1. After the coin is flipped 5 times, what is the probability that Kate has more than$10 but less than $15? 5/16 15/32 Â½ 21/32 11/16 As noted above, Kate will have between$10 and $15 if the coin lands on tails 3 or 4 times. 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. Kate already has$10 . So for the probability that at the end he must have more that $10 but less than$ 15 , at least one but not 5 tails should have appeared . So , why aren't we considering 1 and 2 appearances that would in turn earn him 1 or 2 dollars.

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Deepthi Subbu wrote:
GMATGuruNY wrote:
rtaha2412 wrote:
Kate and David each have $10. Together they flip a coin 5 times. Every time the coin lands on heads, Kate gives David$1. Every time the coin lands on tails, David gives Kate $1. After the coin is flipped 5 times, what is the probability that Kate has more than$10 but less than $15? 5/16 15/32 Â½ 21/32 11/16 As noted above, Kate will have between$10 and $15 if the coin lands on tails 3 or 4 times. 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. Kate already has$10 . So for the probability that at the end he must have more that $10 but less than$ 15 , at least one but not 5 tails should have appeared . So , why aren't we considering 1 and 2 appearances that would in turn earn him 1 or 2 dollars.
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.

_________________
Mitch Hunt
Private Tutor for the GMAT and GRE
GMATGuruNY@gmail.com

If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon.

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Master | Next Rank: 500 Posts
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GMATGuruNY wrote:
Deepthi Subbu wrote:
GMATGuruNY wrote:
rtaha2412 wrote:
Kate and David each have $10. Together they flip a coin 5 times. Every time the coin lands on heads, Kate gives David$1. Every time the coin lands on tails, David gives Kate $1. After the coin is flipped 5 times, what is the probability that Kate has more than$10 but less than $15? 5/16 15/32 Â½ 21/32 11/16 As noted above, Kate will have between$10 and $15 if the coin lands on tails 3 or 4 times. 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. Kate already has$10 . So for the probability that at the end he must have more that $10 but less than$ 15 , at least one but not 5 tails should have appeared . So , why aren't we considering 1 and 2 appearances that would in turn earn him 1 or 2 dollars.
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.
Oh ya , got it

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GMATGuruNY wrote:
rtaha2412 wrote:
Kate and David each have $10. Together they flip a coin 5 times. Every time the coin lands on heads, Kate gives David$1. Every time the coin lands on tails, David gives Kate $1. After the coin is flipped 5 times, what is the probability that Kate has more than$10 but less than $15? 5/16 15/32 Â½ 21/32 11/16 As noted above, Kate will have between$10 and $15 if the coin lands on tails 3 or 4 times. 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. Dear GMATGuru I encountered another solution in which total possibility = 2^5 (which will include a certain number of similar outcomes in different orders) Since it does not matter which of the 5 tosses is won, as long as it's either 3 or 4 of them, we can use the combination formula: For 3 wins, we have 5!/3!2! = 10 possible combinations of 3 wins For 4 wins, we have 5!/4!1! = 5 possible combinations of 4 wins My question here: How we can divide combinations over permutation? as As far as I know, we must either deal with all combinations (in numerator and denominator) or all permutations (in numerator and denominator). Should not we discard similar outcome from 2^5 to make numerator matches the denominator? or the other way to consider the order in the numerator to match the order in the denominator? ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15387 messages Followed by: 1872 members Upvotes: 13060 GMAT Score: 790 Mo2men wrote: [Dear GMATGuru I encountered another solution in which total possibility = 2^5 (which will include a certain number of similar outcomes in different orders) Since it does not matter which of the 5 tosses is won, as long as it's either 3 or 4 of them, we can use the combination formula: For 3 wins, we have 5!/3!2! = 10 possible combinations of 3 wins For 4 wins, we have 5!/4!1! = 5 possible combinations of 4 wins My question here: How we can divide combinations over permutation? as As far as I know, we must either deal with all combinations (in numerator and denominator) or all permutations (in numerator and denominator). 3 wins and 2 losses = WWWLL. One way to determine the number of ways to get 3 wins is to count the number of ways to arrange the 5 letters WWWLL. The number of ways to arrange 5 elements = 5!. But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED. The reason: When the identical elements swap positions, the arrangement doesn't change. Here, we must divide by 3! to account for the 3 identical W's and by 2! to account for the 2 identical L's: 5!/(3!2!) = 10 Another approach: There are 5 positions in the arrangement. Each position must be occupied by W or L. A good arrangement occurs when W's occupy A COMBINATION OF 3 POSITIONS in the 5-position arrangement. Thus, to determine the number of ways to get 3 wins, we can count the number ways to choose a combination of 3 positions from 5 options: 5C3 = 5!/(3!2!) = 10 Each approach yields the same result. _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • Free Practice Test & Review How would you score if you took the GMAT Available with Beat the GMAT members only code • 5-Day Free Trial 5-day free, full-access trial TTP Quant Available with Beat the GMAT members only code • Free Trial & Practice Exam BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200

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