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Updated on: 17 Feb 2015, 12:34
1
13
00:00

Difficulty:

55% (hard)

Question Stats:

68% (02:42) correct 32% (02:32) wrong based on 218 sessions

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30 Oct 2014, 09:55
3
8
Senior Manager
Affiliations: SPG
Joined: 15 Nov 2006
Posts: 296

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22 Sep 2009, 14:40
1
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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19 Oct 2011, 11:20
4
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 !! Senior Manager Joined: 13 Jun 2013 Posts: 269 Re: Kate and Danny each have$10. Together, they flip a fair coin 5 times.  [#permalink]

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31 Oct 2014, 02:56
7
1
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 Intern Joined: 07 May 2015 Posts: 16 Location: India Concentration: International Business, Finance GPA: 3.5 WE: Information Technology (Computer Software) Re: Kate and Danny each have$10. Together, they flip a fair coin 5 times.  [#permalink]

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29 Jun 2015, 07:45
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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Joined: 31 May 2017
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30 Jul 2018, 22:15
My explanation is
Total outcomes of tossing a coin =2^5 =32

So the probability is exactly half times head and half time tails...i.e.16/32
And since it is asked for 10-15

Is my approach right??

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Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink] ### Show Tags 06 Feb 2019, 19:32 powerka wrote: 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

In order for Kate to have more than 10 dollars but less than 15 dollars, either of the following two outcomes must have occurred:

T-T-T-T-H, so Kate would have 13 dollars

T-T-T-H-H, so Kate would have 11 dollars

Let’s calculate the probability of each outcome:

P(T-T-T-T-H) = (1/2)^5 = 1/32

Since T-T-T-T-H can be arranged in 5!/4! = 5 ways, the probability is 5/32.

Next:

P(T-T-T-H-H) = (1/2)^5 = 1/32

Since T-T-T-H-H can be arranged in 5!/(3! x 2!) = 10 ways, the probability is 10/32.

So the overall probability that Kate has more than $10 but less than$15 is 15/32.

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