Question

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

Answer 1

Answer:

the probability that exact three would be replaced under each warranty is 0.34

Step-by-step explanation:

The computation of the probability that exact three would be replaced under each warranty is as follows

Given that

Submitted under warranty = P = 20%

Replaced with the new units = 40%

So replaced and submitted would be

= 40% × 20%

= 80%

Now let us assume the number of telephones be x for ending up for replacement

Total number of telephones purchased is 10

Now the probability by applying the formula of binomial probability is

$P(X) = n_C_xp^x(1 - p)^{n-x}$

$P (X = 3) = ^{10}C_3(0.08)^3 (1 - 0.08)^{10 - 3}\\\\= \frac{10!}{3!(10! - 3!)} (0.08)^3(0.92)^7$

= 0.34

Hence, the probability that exact three would be replaced under each warranty is 0.34