Posted by Trisha on Tuesday, July 31, 2012 at 2:01am.
If we can assume random trials, and that probability remains constant (at 1/2).
Use binomial distribution.
N=5 (no. of trials),
n=3 (no. of successes),
p=1/2 (probability), q=(1-p)=1/2
P(3 out of 5)
=C(5,3)p^3q^(5-3)
=5!/(2!3!) (1/2)^3(1/2)^2
=10/32
=5/16
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