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by Nidhs » Fri Feb 08, 2008 6:02 pm
Five fair coins are tossed. What is the probability that exactly three of the coins land tails side up?

a) 5/32
b) 3/16
c) 5/16
d) 3/8
e) 5/8


The OA is c but i don't know how to solve it.
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by gmatguy16 » Sat Feb 09, 2008 9:44 am
since only heads or tails is possible we can use binomial distribution ..
answer is 5 c 3 (0.5) ^ 3 (0.5) ^ (5-3) = 10(0.5) ^ 3 (0.5 )^ 2 = 5/16
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by Stuart@KaplanGMAT » Sat Feb 09, 2008 11:18 am
gmatguy16 wrote:since only heads or tails is possible we can use binomial distribution ..
answer is 5 c 3 (0.5) ^ 3 (0.5) ^ (5-3) = 10(0.5) ^ 3 (0.5 )^ 2 = 5/16
For coin flips we know that the number of possible outcomes is always 2^n, where n is the number of flips.

So, the generic formula is:

nCk/2^n

where n is the number of flips and k is the number of results you want to get.

In this question, we have n=5 and k=3, so we'd have:

5C3/2^5
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by xilef » Thu Feb 14, 2008 5:25 pm
Another way of looking at it is:

P of getting exactly 3 of five is 1/2*1/2*1/2*1/2*1/2 = 1/32

and there are 10 ways of getting 3 in 5 tosses, so:

1/32*10=5/16
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