# Binomial probability help

A fair coin is tossed 5 times. What is the probability of obtaining exactly 2 heads if it is known that at least 1 head appeared?

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mathman

The way I see it, this is a conditional probability problem that involves the binomial distribution. That is, "what is the probability of getting exactly 2 heads given that at least 1 head occurs".

The formula for the probabilty of event A given B, written as P(A|B), is:
P(A|B) = $\frac{P(A\bigcap B)}{P(B)}$.

In this case, P(A) = probability of exactly 2 heads, and P(B) = probability of ≥ 1 head.
Since P(A) fulfills P(B) as well, P(A$\bigcap$B) = P(A)= (10 choose 2)(0.5^5)
= 0.3125.

P(B) is a bit trickier. It's basically the P(1 H) + P(2 H) + P(3 H) + P(4 H) + P(5 H), which can also be written as 1 - P(0 heads) = 1 - (0.5)^5 = 0.96875.

Therefore, P(A|B) = 0.3125 / 0.96875 ≈ 0.323

EDIT - this is basically the full solution to what mathman just said.

thanks pshooter that explained so much better!