Let p be the probability of a ATM breakdown,
p=0.2
i. there are no breakdowns,
P1=(1−p)5P1=(0.8)5P1=0.32768
ii. four machines break down
P2=(p4)×(1−p)P2=(0.2)4×(0.8)P2=0.00128
iii. all machines break down
P3=p5P3=(0.2)5P3=0.00032
iv. at least three machines break down
P4=P2+P3+(p3×(1−p)2)P4=0.00128+0.00032+0.23×0.82P4=0.00672
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