Question

Thirty percent of all telephones of a certain type are submitted for service while under warranty. Of these, 70% can be repaired, whereas the other 30% must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly two will end up being replaced under warranty? (Round your answer to three decimal places.)

Answer 1

Answer:

$X \sim Binom(n=10, p=0.3)$

And for this case we want to find this probability:

$P(X =2)$

And we can use the probability mass function and we got:

$P(X=2)=(10C2)(0.3)^2 (1-0.3)^{10-2}=0.233$

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

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".

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:  

$nCx=\frac{n!}{(n-x)! x!}$  

Solution to the problem

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

$X \sim Binom(n=10, p=0.3)$

And for this case we want to find this probability:

$P(X =2)$

And we can use the probability mass function and we got:

$P(X=2)=(10C2)(0.3)^2 (1-0.3)^{10-2}=0.233$