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November 29, 2014

November 29, 2014

Posted by **Trisha** on Tuesday, July 31, 2012 at 2:01am.

- Math -
**MathMate**, Tuesday, July 31, 2012 at 7:10pmIf 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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