a.
cumulative distribution function:
F(x)=P(X≤k)=i=0∑k(ni)pi(1−p)n−i
where k is successes in n independent Bernoulli trials
p is probability of success in trial
we have:
n=5,p=0.3
then:
F(x)=P(X≤k)=i=0∑k(5i)0.3i0.75−i
b.
g(0)=F(0)=P(X≤0)=0.75=0.1681
g(1)=F(1)=P(X≤1)=P(0)+P(1)
P(1)=5⋅0.3⋅0.74=0.3602
g(1)=0.1681+0.3602=0.5283
g(2)=F(2)=P(X≤2)=P(0)+P(1)+P(2)
P(2)=C52⋅0.32⋅0.53=0.1125
g(2)=0.1681+0.3602+0.1125=0.6408
g(3)=F(3)=P(X≤3)=g(2)+P(3)
P(3)=C53⋅0.33⋅0.52=0.0675
g(3)=0.6408+0.0675=0.6583
g(4)=F(4)=P(X≤4)=g(3)+P(4)
P(4)=C54⋅0.34⋅0.5=0.0203
g(4)=0.6583+0.0203=0.6786
g(5)=F(5)=P(X≤5)=1
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