Question

Each year a company selects a number of employees for a management training program. On average, 30 percent of those sent complete the program. Out of the 14 people sent, what is the probability that exactly 5 complete the program?

a) 0.1963
b) 0.3963
c) 0.8805
d) 0.7805
e) 0.2403
f) None of the above

Answer 1

`P(X=5)=(14C5)(0.3)^5 (1-0.3)^(14-5)=0.1963`

The correct answer would be:

a) 0.1963

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

Solution to the problem

Let X the random variable of interest, "number of people who complete the program" on this case we now that:

`X \sim Binom(n=14, p=0.3)`

And we want to find this probability:

`P(X=5)`

Using the probability mass function we got:

`P(X=5)=(14C5)(0.3)^5 (1-0.3)^(14-5)=0.1963`

The correct answer would be:

a) 0.1963