Question

Thirty 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 four will end up being replaced under warranty? (Round your answer to three decimal places.)

Answer 1

Answer:

The probability that exactly four will end up being replaced under warranty

P(X=4) = 0.20012

Step-by-step explanation:

Explanation:-

by using binomial distribution

P(X=r) = $n_{Cr } p^{r} q^{n-r}$

Given data Thirty percent of all telephones of a certain type are submitted for service while under warranty.

we can take P = 30/100 = 0.3

q = 1-p = 1-0.3 = 0.7

 If a company purchases ten of these telephones that is n= 10

The probability that exactly four will end up being replaced under warranty.

$P(X=4) = 10_{C4 } (0.3)^{4} (0.7)^{10-4}$

we will use formula

$n_{Cr } = \frac{n!}{(n-r)!r!}$

$10_{C4 } = \frac{10!}{(10-4)!4!}$          = 120

$P(X=4) = 120(0.3)^{4} (0.7)^{10-4}$

on simplification we get P(x=4) = 0.2001

The probability that exactly four will end up being replaced under warranty.

P(x=4) = 0.20012