Unfair Coin TossingSolutionPH 04 a the probability that at

Unfair Coin Tossing...

Solution

P(H) = 0.4

a) the probability that at least one head occurs = 1 - probability of no head occurs

= 1 - (0.6)*(0.6) = 1-0.36

=0.64

b)

P(X=0,Y=0) = 0.6*0.6 = .36

P(X=1,Y=0) = 0 (Since, this case is not possible)

P(X=0,Y=1) = 0.6*0.4 = 0.24

P(X=1,Y=1) = 0.4*0.6 = 0.24

P(X=0,Y=2) = 0 (Since, this case is not possible)

P(X=1,Y=2) = 0.4*0.4 = 0.16