Question

An important part of customer service responsibilities of a cable company relates to the speed with which trouble in service can be repaired. Historically, the data show that the likelihood is 0.75 that troubles in a residential service can be repaired on the same day. For the first 5 troubles reported on a given day, what is the probability that fewer than 2 troubles will be repaired on the same day?

Answer 1

The answer to your question is
D.  .2373


Hope this helps

Answer 2

Answer:

The probability that fewer than 2 troubles will be repaired on the same day is:

0.015576

Step-by-step explanation:

We have to use the Binomial function is n=order to calculate the probability of r successes out of the n-events.

We know that the probability of r success out of the n-events is given by:

$P(X=r\ successes)=n_C_r\cdot p^r\cdot (1-p)^{n-r}$

where p is the probability of an success.

Here we have:

n=5 and r=0,1.

since we are asked to find the probability that fewer than 2 troubles will be repaired on the same day.

i.e we have to find:

$P(X$

Also p=0.75 ( since, likelihood is 0.75 that troubles in a residential service can be repaired on the same day )

Hence,

$P(X=0)=5_C_0\cdot (0.75)^{0}\cdot (1-0.75)^{5-0}\\\\P(X=0)=1\cdot 1\cdot (0.25)^5\\\\P(X=0)=0.000976$

and

$P(X=1)=5_C_1\cdot (0.75)^{1}\cdot (0.25)^{5-1}\\\\P(X=1)=5\cdot (0.75)\cdot (0.25)^4\\\\P(X=1)=0.0146$

Hence,

$P(X$

Hence,  the probability that fewer than 2 troubles will be repaired on the same day is:

0.015576