Question

a) here as number of trails are independent and probability of head on each event is fixed ; therefore it is binomial distribution with parameter n=5 and p=0.5

hence probability mass function of X:

P(X=x)=

frm above P(X=0)= =0.03125

P(X=1)= =0.15625

P(X=2)=0.3125

P(X=3)=0.3125

P(X=4)=0.15625

P(X=5)=0.03125

probbaility distribution function F(x) of X is as follows:

 0 x<=0 0.03125 0<=x<1 0.1875 1<=x<2 F(x)= 0.5 2<=x<3 0.8125 3<=x<4 0.96875 4<=x<5 1 x>=5

b)

P(realizing at least 3 heads out of 5) =P(X=3)+P(X=4)+P(X=5) =0.3125+0.15625+0.03125=0.5

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