Question:If a fair coin is tossed at random five independent times, find the con

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Question:If a fair coin is tossed at random five independent times, find the conditional probability of five heads, given that there are at least four heads.

Book's answer: 1/6

My approach:
H=heads
T=tails
conditional probability: p(A|B)=[p(A)np(B)]/p(B)
since they are independent above equation reduces to [p(A)*p(B)]/p(B)

p(5H|x=4H)
=p(5H|1-p(x=3H))
=[p(5H)n(1-p(x=3H))]/p(1-p(x=3H)

I looked up the values of p(5H) and p(x=3H) in the binomial distribution table this is what I got for b(5,.5)
p(5H)=.0312
p(x=3H)=.1875

When I plugged in these numbers in the above equation I got:
[(.0312)(.1875)]/(.1875)=.0312?1/6

Can anyone please help enlighten me on how to get the correct answer.

Explanation / Answer

Ok: So we know there are at least 4 heads, meaning there are either 4 or 5 heads. So we want there to be 4. Probability of 5 heads/probability of 4+5 heads (since it's an or, we add) HHHHT HHHTH HHTHH HTHHH and THHHH 5 possibilities, while only one HHHHH for 5 heads. 1/6. If it doesn't make sense, mail me.

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