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# Probability Problem - Kate and Danny each have $10.  Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: Manager Joined: 22 Jul 2009 Posts: 192 Followers: 4 Kudos [?]: 185 [0], given: 18 Probability Problem - Kate and Danny each have$10. [#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: D _________________ Please kudos if my post helps.  Manhattan GMAT Discount Codes Kaplan GMAT Prep Discount Codes Veritas Prep GMAT Discount Codes Manager Joined: 22 Jul 2009 Posts: 192 Followers: 4 Kudos [?]: 185 [0], given: 18 Re: Probability Problem - Kate and Danny each have$10. [#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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Please kudos if my post helps.

Manager
Joined: 04 Sep 2009
Posts: 53
WE 1: Real estate investment consulting
Followers: 1

Kudos [?]: 16 [0], given: 7

Re: Probability Problem - Kate and Danny each have $10. [#permalink] 22 Sep 2009, 14:15 Came there same way. Made a mistake 1st (already half a sleep for an excuse ) and calculated only 4T1H. Senior Manager Affiliations: SPG Joined: 15 Nov 2006 Posts: 326 Followers: 11 Kudos [?]: 242 [2] , given: 20 Re: Probability Problem - Kate and Danny each have$10. [#permalink]  01 Mar 2011, 04:28
2
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}

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press kudos, if you like the explanation, appreciate the effort or encourage people to respond.

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: Probability Problem - Kate and Danny each have $10. [#permalink] 19 Oct 2011, 08:26 y arent we taking into account the probablility of 2 heads or 1 Manager Status: Next engagement on Nov-19-2011 Joined: 12 Jan 2011 Posts: 84 Location: New Delhi, India Schools: IIM, ISB, & XLRI WE 1: B.Tech (Information Technology) Followers: 0 Kudos [?]: 6 [0], given: 5 Re: Probability Problem - Kate and Danny each have$10. [#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 _________________ GMAT is an addiction and I am darn addicted Preparation for final battel: GMAT PREP-1 750 Q50 V41 - Oct 16 2011 GMAT PREP-2 710 Q50 V36 - Oct 22 2011 ==> Scored 50 in Quant second time in a row MGMAT---- -1 560 Q28 V39 - Oct 29 2011 ==> Left Quant half done and continued with Verbal. Happy to see Q39 My ongoing plan: http://gmatclub.com/forum/550-to-630-need-more-to-achieve-my-dream-121613.html#p989311 Appreciate by kudos !! Re: Probability Problem - Kate and Danny each have$10.   [#permalink] 19 Oct 2011, 10:20
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