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Re: Kevin flips a coin four times. WHat is the probability the he gets hea [#permalink]
Getting Head on at least one flip of four= P(1 Head) + P(2 Head) + P(3 Head) + P(4 Head)

Also, P (0 Head) + P(1 Head) + P(2 Head) + P(3 Head) + P(4 Head) = 1

Therefore, Getting Head on at least one flip of four= P(1 Head) + P(2 Head) + P(3 Head) + P(4 Head) = 1- P (0 Head)

P (0 Head)= (1/2)*(1/2)*(1/2)*(1/2) = (1/16)

Probability of getting Head on at least one flip of four =1- P (0 Head) = 1- (1/16) = (15/16)

IMO, Answer is E
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Kevin flips a coin four times. WHat is the probability the he gets hea [#permalink]
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P(0H) + P(1H) + P(2H) + P(3H) + P(4H) = 1
=> P(1H) + P(2H) + P(3H) + P(4H) = 1 - P(0H)

P(At least 1 H) = 1 - P(0H)

1 - P(4T) = 1 - \(\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}\) = 1 - \(\frac{1}{16}\) = \(\frac{15}{16}\)

So, Answer will be E
Hope it helps!

Watch the following video to learn How to Solve Probability with Coin Toss Problems

GMAT Club Bot
Kevin flips a coin four times. WHat is the probability the he gets hea [#permalink]
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Math Expert
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