Conditional Probability with coins.

Fair coin is flipped five independent, random times. Calculate the conditional probability of 5 heads, knowing that there were at least 4 heads.

I came up with P(five heads) = 1 - P(four heads ) = 1 - (5C4)/2^5 = 27/32. This is wrong since I KNOW the answer is 1/6. I believe this is the correct approach, however. Please help me find how to get that solution. Thanks.

nevermind, i just figured out =)

Fair coin is flipped five independent, random times. Calculate the conditional probability of 5 heads, knowing that there were at least 4 heads.

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We want \(\displaystyle P(\text{5 heads}|\text{at least 4 heads})\)

The probability of 5 heads in 5 tosses is \(\displaystyle \left(\frac{1}{2}\right)^{5}=\frac{1}{32}\)

The probability of at least 4 heads in 5 tosses is \(\displaystyle \sum_{k=4}^{5}\binom{5}{k}(\frac{1}{2})^{k}(\frac{1}{2})^{5-k}=\frac{3}{16}\)

\(\displaystyle \frac{\frac{1}{32}}{\frac{3}{16}}=\frac{1}{6}\)

*************EDIT: Oops, I was typing while you posted.

The help is still appreciated. Thanks!

in conditional probablity the formula is P(A|B)=P(A/\B)/P(B)
but here you have done P(A)/P(B).