Partial solutions to 2004 Exam 2.

(a) Y ~ Bin(7,0.25).
    mu = 7(0.25) = 1.75.
    sigma = sqrt(7(0.25)(0.75)) = 1.15

(b) Y ~ Bin(7,0.5)
    P(Y=0) + P(Y=1) + P(Y=2)
    = 0.0078 + 0.0547 + 0.1641
    = 0.2266

(c) Y ~ Bin(7,0.25)
    P(Y=2) = 0.3115

(d) Law of total probability
    P(Y=2) = 0.36(0) + 0.48(0.3115) + 0.16(0.1641)
           = 0.1758

(e) Bayes' Theorem
    P(Aa | Y=2) = P(Aa & Y=2) / P(Y=2)
                = 0.48(0.3115) / 0.1758
                = 0.8505