Posted by Mujtaba on Saturday, January 5, 2013 at 3:39am.
I assume that each of the sixteen events has two outcomes of S, namely
P(s1,s1), P(s1,s2).....P(s4,s4).
By the multiplication rule, and assuming that the two outcomes are independent, we have
P(s1,s1)=P(s1)P(s1)=0.5*0.5;=0.25
....
P(s4,s4)=P(s4)P(s4)=0.125*0.125=0.015625
Note: The sum of probabilities of the 16 events should add up to 1.
Not sure about the surprise part.
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