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# Kevin flips a coin four times. what is the probability that

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Kevin flips a coin four times. what is the probability that [#permalink]  05 Nov 2016, 06:10
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Question Stats:

100% (00:27) correct 0% (00:00) wrong based on 2 sessions
Kevin flips a coin four times. what is the probability that he gets heads on at least one of the four flip?

a) $$\frac{1}{16}$$

b) $$\frac{1}{4}$$

c) $$\frac{3}{4}$$

d) $$\frac{13}{16}$$

e) $$\frac{15}{16.}$$

books says its option 'e' but i say its 'b'

what do you think?

justifiction:

[p(H)*p(T)*p(T)*p(T)] + [p(H)*p(H)*p(T)*p(T)] + [p(H)*p(H)*p(H)*p(T)] + [p(H)*p(H)*p(H)*p(H)]

=> [1/2 * 1/2 *1/2*1/2] + [1/2 * 1/2 *1/2*1/2] + [1/2 * 1/2 *1/2*1/2] + [1/2 * 1/2 *1/2*1/2] = 4/16 => 1/4.

thank you
siva
[Reveal] Spoiler: OA
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Re: what is the answer 1/4 or 15/16 [#permalink]  05 Nov 2016, 08:06
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Expert's post
Hi,

The only case where we do not have a head is the case TTTT.

P(TTTT)=$$\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}$$=$$\frac{1}{16}$$.

So P(At least one head)= $$1 -\frac{1}{16}=\frac{15}{16}$$.

Hence E is correct option.
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Sandy
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Re: what is the answer 1/4 or 15/16   [#permalink] 05 Nov 2016, 08:06
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