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Source: — Problem Solving |

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by Anurag@Gurome » Fri Jul 29, 2011 5:40 am
ruplun wrote:Kevin flips a coin 4 times .What is the probability that he gets heads on at least one of the 4 flips?
Probability that Kevin gets heads on none of the 4 flips = (1/2)^4 = 1/16
Therefore, probability that Kevin gets heads on at least one of the 4 flips = 1 - 1/16 = [spoiler]15/16[/spoiler]
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by Brent@GMATPrepNow » Fri Jul 29, 2011 5:41 am
ruplun wrote:Kevin flips a coin 4 times .What is the probability that he gets heads on at least one of the 4 flips?
We can solve this, we'll using the complement (complements are often useful for questions involving "at least").

So, P(at least 1 head) = 1 - P(no heads)
P(no heads) = P(all tails)

Aside: we'll use T1 to denote tails on first flip, T2 for tails on 2nd flip and so on.

P(all tails) = P(T1 AND T2 AND T3 AND T4)
P(all tails) = P(T1) x P(T2) x P(T3) x P(T4)
P(all tails) = 1/2 x 1/2 x 1/2 x 1/2
P(all tails) = 1/16

So, P(at least 1 head) = 1 - P(no heads)
P(at least 1 head) = 1 - 1/16
P(at least 1 head) = 15/16

Cheers,
Brent
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