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# Kate and Danny each have $10. Together, they flip a fair coin 5 times.  Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: Manager Joined: 22 Jul 2009 Posts: 193 Followers: 4 Kudos [?]: 210 [0], given: 18 Kate and Danny each have$10. Together, they flip a fair coin 5 times. [#permalink]  22 Sep 2009, 13:36
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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 Source: Manhattan GMAT Archive (tough problems set).doc [Reveal] Spoiler: OA _________________ Please kudos if my post helps. Last edited by Bunuel on 17 Feb 2015, 11:34, edited 1 time in total. Renamed the topic, edited the question and added the OA.  Manhattan GMAT Discount Codes Kaplan GMAT Prep Discount Codes GMAT Pill GMAT Discount Codes Manager Joined: 22 Jul 2009 Posts: 193 Followers: 4 Kudos [?]: 210 [0], given: 18 Re: Kate and Danny each have$10. Together, they flip a fair coin 5 times. [#permalink]  22 Sep 2009, 13:40
Decided to post this question as Manhattan's explanation was unnecessarily cumbersome.

Question asks for the probability of the coin landing tails up either 3 or 4 times = P(3t) + P(4t)

Binomial distribution formula: nCk p^k (1-p)^(n-k)

P(3t) = 5C3 (1/2)^3 (1/2)^2 = 10 (1/2)^5
P(4t) = 5C4 (1/2)^4 (1/2)^1 = 5 (1/2)^5

=> P(3t) + P(4t) = 15/32
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Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink] 01 Mar 2011, 04:28 3 This post received KUDOS Q: what is the probability of getting 3 or 4 heads if a fair coin is flipped 5 times. total number of possibilities $$= 2 * 2 * 2 * 2 * 2 = 2^5$$ $$P(3) = \frac{5C3}{2^5} = \frac{10}{32}$$ $$P(4) = \frac{5C4}{2^5} = \frac{5}{32}$$ Required probability $$= P(3) + P(4) = \frac{15}{32}$$ _________________ press kudos, if you like the explanation, appreciate the effort or encourage people to respond. Download the Ultimate SC Flashcards Intern Joined: 17 Oct 2011 Posts: 2 Concentration: Entrepreneurship, Marketing GMAT Date: 11-25-2011 GPA: 3.9 Followers: 0 Kudos [?]: 0 [0], given: 0 Re: Kate and Danny each have$10. Together, they flip a fair coin 5 times. [#permalink]  19 Oct 2011, 08:26
y arent we taking into account the probablility of 2 heads or 1
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Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink] 19 Oct 2011, 10:20 deep5586 wrote: y arent we taking into account the probablility of 2 heads or 1 For 2 & 1 heads Kate will end up with <$10 and we want her to win . Therefore, only possibilities are 3 or 4 heads.

I made an educated guess and it worked fine.

Ans- 'D'

MGMAT's anagram helped here as well.

HHHHT = 5!/4!*1! = 5
HHHTT = 5!/3!*2! = 10

Total acceptable cases = 15
Total cases = 32

P = 15/32
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Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink] 30 Oct 2014, 08:55 Expert's post 2 This post was BOOKMARKED Tough and Tricky questions: Probability. 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
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Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink] 31 Oct 2014, 01:56 2 This post received KUDOS Bunuel wrote: Tough and Tricky questions: Probability. 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

total number of possible outcomes when coin will be flipped 5 times= 2^5=32

now, for kate to have more than $10, she must have min. 3 favorable and max. 4 favorable draws in the 5 throws. (because if she wins in all the 5 draws, then she will have$15)

kate will win, every time tails (T) appears on the coin.

case 1) kate wins in 3 draws and losses in two. so we have 3 T's and 2H's(T,T,T,H,H), which can arrange themselves in 5!/(3!)(2!)=10

case 2) kate wins in 4 draws and losses in the 1 draw. so we have 4 T's and 1H (T,T,T,T,H), which can arrange themselves in 5!/4!=5

thus total no. of favorable ways=10+5=15
and thus required probability=15/32
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Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink] 17 Feb 2015, 11:16 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re: Kate and Danny each have$10. Together, they flip a fair coin 5 times.   [#permalink] 17 Feb 2015, 11:16
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