Question

An appliance repairman services five washing machines on site each day. One-third of the service calls require installation of a particular part. 1. The repairman has only one such part on his truck today. Find the probability that the one part will be enough today, that is, that at most one washing machine he services will require installation of this particular part 2. Find the minimum number of such parts he should take with him each day in order that the probability that he have enough for the day's service calls is at least 95%.

Answer 1

Let X be the random variable number of washing machines he has to install on site each day

Then X ~ Binomial (5,1/3) As each day he services 5 washing machines and probability of a service including installation is 1/3.

1)

Probability that at most one washing machine he services requires installation is

P[X<2] = P[X=0] + P[x=1]

=0.1317 + 0.3292

=0.4609

2)

To find x such that P[X<=x] =0.95

Given n=5, and p=1/3

We get the cumulative distribution table

x	P[X<=x]
0	0.1317
1	0.4609
2	0.7901
3	0.9547
4	0.9959
5	1

Clearly for X=3 cumulative probability is just above 95% (95.47%)

Thus the minimum number of parts that he should take with him in order that the probability hehas enough for the day’s service calls is 3