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## If a certain coin is flipped

tagged by: Brent@GMATPrepNow

This topic has 2 expert replies and 0 member replies

### Top Member

rsarashi Master | Next Rank: 500 Posts
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#### If a certain coin is flipped

Sat Sep 02, 2017 11:42 am
Elapsed Time: 00:00
• Lap #[LAPCOUNT] ([LAPTIME])
If a certain coin is flipped, the probability that the coin will land heads up is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?

A) 3/5

B) 1/2

C) 1/5

D) 1/8

E) 1/32

OAE

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### GMAT/MBA Expert

Brent@GMATPrepNow GMAT Instructor
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Sat Sep 02, 2017 11:49 am
rsarashi wrote:
If a certain coin is flipped, the probability that the coin will land heads up is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?
A) 3/5
B) 1/2
C) 1/5
D) 1/8
E) 1/32

OAE
On ANY individual toss, P(heads) = 1/2 and P(tails) = 1/2

P(heads 1st toss AND heads 2nd toss AND heads 3rd toss AND tails 4th toss AND tails 5th toss AND)
= P(heads 1st toss) x P(heads 2nd toss) x P(heads 3rd toss) x P(tails 4th toss) x P(tails 5th toss)
= 1/2 x 1/2 x 1/2 x 1/2 x 1/2
= 1/32

Cheers,
Brent

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### GMAT/MBA Expert

Jay@ManhattanReview GMAT Instructor
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Sun Sep 03, 2017 3:45 am
rsarashi wrote:
If a certain coin is flipped, the probability that the coin will land heads up is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?

A) 3/5

B) 1/2

C) 1/5

D) 1/8

E) 1/32

OAE
I'm surprised that there is no answer as 5/16. This makes someone safe in case he/she does not get the question well.

Brent explained this one quite well.

Quote:
If a certain coin is flipped, the probability that the coin will land heads up is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on 3 flips and tail on 2 flips?
Unlike the original question in which the condition to get heads on the first 3 flips and tail to get on the last 2 flips, in the above-modified question, there is no such condition, thus, only important is to get 3 head out of 5 flips, be it in any order. A couple of instances HTHTH and HHTTH are valid.

Three instances of heads out of five instances is given by 5C3 = (5.4.3) / (1.2.3) = 10 ways.

Thus, the required probability = 5C3*[(1/2*1/2*1/2)*(1/2*1/2)] = 10*[1/2^5] = 5/16.

Hope this helps!

-Jay
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