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Probability Problem - Kate and Danny each have $10.

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Probability Problem - Kate and Danny each have $10. [#permalink] New post 22 Sep 2009, 14:36
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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

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Re: Probability Problem - Kate and Danny each have $10. [#permalink] New post 22 Sep 2009, 14:40
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: Probability Problem - Kate and Danny each have $10. [#permalink] New post 22 Sep 2009, 15:15
Came there same way. Made a mistake 1st (already half a sleep for an excuse :) ) and calculated only 4T1H.
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Re: Probability Problem - Kate and Danny each have $10. [#permalink] New post 01 Mar 2011, 05:28
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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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Re: Probability Problem - Kate and Danny each have $10. [#permalink] New post 19 Oct 2011, 09:26
y arent we taking into account the probablility of 2 heads or 1
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Re: Probability Problem - Kate and Danny each have $10. [#permalink] New post 19 Oct 2011, 11:20
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. :-D

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 :-D
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Re: Probability Problem - Kate and Danny each have $10.   [#permalink] 19 Oct 2011, 11:20
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