SOLUTION: A balanced coin is tossed 8 times. What is the probability of getting exactly 2 heads given that there were at least 2 heads in the 8 tosses?

Question 257819: A balanced coin is tossed 8 times. What is the probability of getting exactly 2 heads given that there were at least 2 heads in the 8 tosses?
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
Let h=.5 and t=.5, probability of heads or tails.
We need to find a number in the expansion of the binomial (h+t)^8
We are looking for h^2t^6
The coefficient is nCr=8C2=28 Combination
28h^2t^6=28*.25*.0156=.1094 prob of exactly 2 heads.
.
1-t^8-8t^7h=1-.0039-(8*.0078*.5)
=1-.0039-.03125
=.9649 prob of at least 2 heads.
.
.1094/.9649=.1134 probability of getting exactly 2 heads given that there were at least 2 heads in the 8 tosses.
.
Ed
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