Calculating Binomial Probability for Coin Tosses with At Least 1 Head

[mtingt]

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?

 

[mathman]

Prob(exactly 2 heads)/(1 - Prob(0 heads))

 

[PShooter1337]

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) = [itex]\frac{P(A\bigcap B)}{P(B)}[/itex].

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[itex]\bigcap[/itex]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.

 

[mtingt]

thanks pshooter that explained so much better!