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Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of these, 70% can be repaired, whereas the other 30% must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly two will end up being replaced under warranty?

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ajeigbeibraheem

Answer:

The required probability is 0.09875

Step-by-step explanation:

From the given information;

the probability of repairing the telephones = 0.70

the probability of the replaced = 0.30

Suppose we consider  Mto denotes the telephone that is submitted for service while under warranty and must be replaced.

Then;

p = P(S) = P(replaced | submitted) P(submitted)

= 0.30 × 0.20

= 0.06

Now, the probability that exactly two will end up being replaced under warranty given that it assumes  a binomial distribution where n = 10 and p = 0.06

[tex]P(X=2)=\bigg (^{10}_{2}\bigg) 0.06^2(1-0.06)^{10-2}[/tex]

[tex]P(X=2)=\dfrac{10!}{2!(10-2)!}\times 0.06^2\times (0.94)^{8}[/tex]

[tex]\mathbf{P(X=2)=0.09875}[/tex]