Problem 1

D = 'C' ( 0.33)

        (Choice) (0.33)
     /     |     \
a=Head      b=Head  c=Head
0.2      0.6        0.8


P(Choice | Head, a=Head) =

P(Choice | Head, b=Head)

P(Choice | Head, c=Head)

Which is maximum of the three?


P(D |Head, a=Head) = P(Head |D) * P(a=Head/D)

Assuming you flip a coin, and it came up as Heads. Which coin is most likely to have produced that result. The answer’s pretty obvious, so provide a probabilistic justification for why you think so.

—

Bayes Rule

P(Choice=c | c=Head) =  P(c=Head|Choice=c)  * P(Choice=c)
                    = 0.8 * 0.3 * 0.3
                    =  0.072

P(Choice=b)                 = 0.6 * 0.3 * 0.3 =  0.054

P(Choice=a)         = 0.2 * 0.3 * 0.3  = 0.018


Max of these. Choice = c.

—

P(C|c=Head) = P(c=Head|C) * P(C)

—

Independent Probability Multiplication

P(HHT|C=a) = P(a=Head|C=a) * P(a=Head|C=a)  * P(a=Tail|C=a)
            = (0.2 * 0.33)  * (0.2 * 0.33) * (0.8 * 0.33)
            = 0.0011499840000000002

            = (0.2 * 0.2 * 0.8)

P(HHT|C=b) = P(b=Head|C=b) * P(b=Head|C=b)  * P(b=Tail|C=b)
           = (0.6 * 0.33) * (0.6 * 0.33) * (0.4 * 0.33)
           = 0.0051749280000000005
           // Winner

           = (0.6 * 0.6 * 0.4)


P(HHT|C=c) = P(c=Head|C=c) * P(c=Head|C=c)  * P(c=Tail|C=c )
           = (0.8 * 0.33) * (0.8 * 0.33) * (0.2 * 0.33)
           = 0.004599936000000001

           = (0.8 * 0.8 * 0.2)