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### Show Tags

Updated on: 17 Feb 2015, 12:34
1
7
00:00

Difficulty:

55% (hard)

Question Stats:

73% (01:48) correct 27% (02:10) wrong based on 246 sessions

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01 Mar 2011, 05:28
10
1
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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##### General Discussion
Manager
Joined: 22 Jul 2009
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19 Oct 2011, 09:26
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: 70
Location: New Delhi, India
Schools: IIM, ISB, & XLRI
WE 1: B.Tech (Information Technology)
Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink] ### Show Tags 19 Oct 2011, 11:20 3 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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31 Oct 2014, 02:56
5
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: 17 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
Intern
Joined: 31 May 2017
Posts: 2

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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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