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probability - coin flips

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by Brent@GMATPrepNow » Wed Dec 17, 2008 10:21 pm
I'd place this one in the 650 range (700+ questions are very hard to compose, but I have a few in the works)

A coins is flipped 6 times. What is the probability that there are exactly 3 heads?

A)1/4
B) 1/3
C)5/16
D) 31/64
E) 1/2

Answer: C
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by cramya » Wed Dec 17, 2008 10:34 pm
Actaully it might be. But due to Stuart's coin flip strategy I was able to solve it

https://www.beatthegmat.com/coin-flip-qu ... 17911.html(post well worth visiting)

Using Pascals triangle for 6 flips

1 6 15 20 15 6 1

The entries in the row represent the different ways to get 0, 1, 2, 3,4,5,6 results respectively

Prob = 20/64 (sum of all the numbers above)
= 5/16
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by scoobydooby » Wed Dec 17, 2008 10:42 pm
total possible outcomes: 2^6 (each flip, 2 outcomes)
3 heads out of 6 flips possible in 6C3 ways or
Probability of 3 heads=6C3/2^6=20/64=5/16
hence C
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by vittalgmat » Wed Dec 17, 2008 10:53 pm
Yep u r right Cramya,
Stuart's techinique help me as well.
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by schumi_gmat » Thu Dec 18, 2008 3:24 pm
There is also a Theorem to solve this problem.

If the event occurs n times and succeeds r times and if p is probability of success and q is probability of failure then, the probability that the event occurs exactly r times is give by -

P(r) = nCr (p)^r (q)n-r

so substuting - P(r=3) = 6C3 (1/2)^3 (1/2)^(6-3)
= 6C3/ 2^6
= 5/16
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