Question

The service manager for a new appliances store reviewed sales records of the past 20 sales of new microwaves to determine the number of warranty repairs he will be called on to perform in the next 90 days. Corporate reports indicate that the probability any one of their new microwaves needs a warranty repair in the first 90 days is 0.05. The manager assumes that calls for warranty repair are independent of one another and is interested in predicting the number of warranty repairs he will be called on to perform in the next 90 days for this batch of 20 new microwaves sold.


What is the probability that less than two of the 20 new microwaves sold will require a warranty repair in the first 90 days?

Answer 1

Solution:

Given that,

p=0.05

and

q=1-p=1-0.05

=0.95

n=20

Now,

P(X<2)=P(X=0)+P(X=1)

$=^{20}C_0\times 0.05^2\times 0.95^{20}+^{20}C_1\times 0.05^1\times 0.95^{19}$

=0.3584+0.3774

=0.7358

Thus, the probability that less than two of the 20 new microwaves sold will require a warranty repair in the first 90 days is 0.7358