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# What is the probability of getting exactly three heads on

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Retired Moderator
Joined: 17 Sep 2013
Posts: 373
Concentration: Strategy, General Management
GMAT 1: 730 Q51 V38
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21 Apr 2014, 21:52
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25% (medium)

Question Stats:

74% (00:54) correct 26% (01:04) wrong based on 449 sessions

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What is the probability of getting exactly three heads on five flips of a fair coin?

(A) 1/32
(B) 3/32
(C) 1/4
(D) 5/16
(E) 1/2

I am looking for a method better than counting...Consider if the Q had 10 coin flips or so..i.e counting would not be a feasible or practical option
[Reveal] Spoiler: OA

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Manager
Joined: 12 May 2013
Posts: 74
Re: What is the probability of getting exactly three heads on [#permalink]

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21 Apr 2014, 23:59
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JusTLucK04 wrote:
What is the probability of getting exactly three heads on five flips of a fair coin?

(A) 1/32
(B) 3/32
(C) 1/4
(D) 5/16
(E) 1/2

I am looking for a method better than counting...Consider if the Q had 10 coin flips or so..i.e counting would not be a feasible or practical option

i use this method, hope it helps
5!/3!*2!*1/2^5 = 5/16 (D)

5!/3!*2! = 5! as we flip the coin 5 times, divided by 3! as we want heads 3 times & 2! for tails

1/2^5 as probability of getting either a heads or a tails is 1/2 raise to 5 because we flip the coin 5 times

please give me kudos if it helps
Math Expert
Joined: 02 Sep 2009
Posts: 45201
Re: What is the probability of getting exactly three heads on [#permalink]

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22 Apr 2014, 01:38
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JusTLucK04 wrote:
What is the probability of getting exactly three heads on five flips of a fair coin?

(A) 1/32
(B) 3/32
(C) 1/4
(D) 5/16
(E) 1/2

I am looking for a method better than counting...Consider if the Q had 10 coin flips or so..i.e counting would not be a feasible or practical option

$$P(HHHTT)=\frac{5!}{3!2!}*(\frac{1}{2})^3*(\frac{1}{2})^2=\frac{10}{32}=\frac{5}{16}$$, we need to multiply by $$\frac{5!}{3!2!}$$ because HHHTT outcome can occur in several ways: HHHTTT, HHTHT, HTHHT, ..., TTHHH ($$\frac{5!}{3!2!}$$ is permutation of 5 letters HHHTT).

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Re: What is the probability of getting exactly three heads on [#permalink]

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03 Sep 2016, 08:01
JusTLucK04 wrote:
What is the probability of getting exactly three heads on five flips of a fair coin?

(A) 1/32
(B) 3/32
(C) 1/4
(D) 5/16
(E) 1/2

I am looking for a method better than counting...Consider if the Q had 10 coin flips or so..i.e counting would not be a feasible or practical option

5 Flips of a fair coin to get = HHHTT = no. of ways this can be achieved = 5!/3!x2! = 10

Probability to get any of the above 10 arrangements (HHHTT) = (1/2)^5 = 1/32

Total probability = 1/32 x 10 = 5/16
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Status: preparing
Joined: 30 Dec 2013
Posts: 41
Location: United Arab Emirates
Concentration: Technology, Entrepreneurship
GMAT 1: 660 Q45 V35
GMAT 2: 640 Q49 V28
GMAT 3: 640 Q49 V28
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GMAT 5: 640 Q49 V28
GPA: 2.84
WE: General Management (Consumer Products)
Re: What is the probability of getting exactly three heads on [#permalink]

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20 Aug 2017, 04:37
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Total number of ways to get 3 heads and 2 tails ( p(h) =p(t)= 1/2 for each)
=2^5=32

# of ways 3 heads can be arranged in 5 tosses= 5c3=10
10/32=5/16 ANS
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Joined: 12 Dec 2016
Posts: 1906
Location: United States
GMAT 1: 700 Q49 V33
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Re: What is the probability of getting exactly three heads on [#permalink]

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25 Aug 2017, 18:14
2 ways:
1/ 5C2 / 2^5 => there are totally 2^5 different results, but only 5C2 favor results,
1/ (1/8) * (1/4) * 5C2
Intern
Joined: 19 Dec 2017
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Re: What is the probability of getting exactly three heads on [#permalink]

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31 Jan 2018, 00:36
HHHTT in any sequence would be required
Probability would be 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 5C3 = 1/32 x 10 = 5/16 (03 heads could be in any position, so 5C3)
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Re: What is the probability of getting exactly three heads on [#permalink]

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08 Feb 2018, 02:39
JusTLucK04 wrote:
What is the probability of getting exactly three heads on five flips of a fair coin?

(A) 1/32
(B) 3/32
(C) 1/4
(D) 5/16
(E) 1/2

I am looking for a method better than counting...Consider if the Q had 10 coin flips or so..i.e counting would not be a feasible or practical option

No. of ways of selecting 3 Heads = $$5C3$$
Total number = $$32$$
$$P = \frac{5C3}{32}$$
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Re: What is the probability of getting exactly three heads on   [#permalink] 08 Feb 2018, 02:39
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