Question

An important part of the 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 five troubles reported on a given day, what is the probability that: Fewer than two troubles will be repaired on the same day? (Points : 3).6328.0010.0156.0146

Answer 1

`P(X < 2)=P(X \leq 1)=P(X=0)+P(X=1)=0.000977+0.0146=0.0156`

Explanation:

1) Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

2) Solution to the problem

Let X the random variable of interest, on this case we now that:

`X \sim Binom(n=5, p=0.75)`

The probability mass function for the Binomial distribution is given as:

`P(X)=(nCx)(p)^x (1-p)^(n-x)`

Where (nCx) means combinatory and it's given by this formula:

`nCx=(n!)/((n-x)! x!)`

And we want to find this probability:

`P(X < 2)=P(X \leq 1)=P(X=0)+P(X=1)`

`P(X=0)=(5C0)(0.75)^0 (1-0.75)^(5-0)=0.000977`

`P(X=1)=(5C1)(0.75)^1 (1-0.75)^(5-1)=0.0146`

`P(X < 2)=P(X \leq 1)=P(X=0)+P(X=1)=0.000977+0.0146=0.0156`