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Question

In a workshop, there are five machines and the probability of any one of them to be out of service on a day is 14. If the probability that at most two machines will be out of service on the same day is (34)3k, then k is equal to :

A
172
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B
4
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C
174
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D
178
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Solution

The correct option is D 178
P(machine being faulty) =p=14
P(machine not being faulty)=q=1p=34
P(at most two machines being faulty)
= P(zero machine being faulty) +
P(one machine being faulty) +
P(two machines being faulty)

= 5C0p0q5+ 5C1p1q4+ 5C2p2q3
=q5+5pq4+10p2q3
=(34)5+5×14(34)4+10×(14)2(34)3
=(34)3[916+1516+1016]
=(34)3×3416=(34)3×178

k=178

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