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Kate and Danny each have $10. Together, they flip a fair coin 5 times.

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Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink]

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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

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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.

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Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink]

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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]

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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: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink]

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New post 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]

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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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Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink]

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Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink]

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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]

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New post 29 Jun 2015, 06:45
Solved it using Pascal's triangle method and able to got the answer in 30 sec. https://en.wikipedia.org/wiki/Pascal's_triangle

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Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink]

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Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times.   [#permalink] 11 Dec 2017, 23:00
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