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Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of these, 60% can be repaired, whereas the other 40% must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly three will end up being replaced under warranty? (Round your answer to three decimal places.)

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JeanaShupp

Answer: 0.034

Step-by-step explanation:

Given : P(Submitted under warranty)= 0.20

P(Replaced  | Submitted under warranty)=0.40

P(Replaced and Submitted under warranty )= P(Submitted under warranty)×P(Replaced  | Submitted under warranty)

=[tex]0.20\times0.40=0.08[/tex]

Let x be the number of telephones will end up being replaced under warranty.

Total telephones purchased : n= 10

Using binomial probability formula : [tex]P(X)=^nC_xp^x(1-p)^{n-x}[/tex]

i.e. The probability that exactly three will end up being replaced under warranty will be :-

[tex]P(X=3)=^{10}C_3(0.08)^3(1-0.08)^{10-3}\\\\=\dfrac{10!}{3!(10-3)!}(0.08)^3(0.92)^7\\\\=0.03427409518\approx0.034[/tex] [Rounded to three decimal places. ]

Hence, the probability that exactly three will end up being replaced under warranty : 0.034

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