Let X= the number of machines which are break down: X∼Bin(n;p)
P(X=x)=(xn)px(1−p)n−x Given that n=5,p=0.2
i. there are no breakdowns
P(X=0)=(05)(0.2)0(1−0.2)5−0==(0.8)5=0.32768
ii. four machines break down
P(X=4)=(45)(0.2)4(1−0.2)5−4==5(0.2)4(0.8)=0.0064 iii. all machines break down
P(X=5)=(55)(0.2)5(1−0.2)5−5==(0.2)5=0.00032 iv. at least three machines break down
P(X≥3)=P(X=3)+P(X=4)+P(X=5)==(35)(0.2)3(1−0.2)5−3+0.0064+0.00032==10(0.2)3(0.8)2+0.00672=0.05792
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