Question

The workers' union at a particular university is quite strong. About 96% of all workers employed by the university belong to the workers' union. Recently, the workers went on strike, and now a local TV station plans to interview workers (chosen at random) at the university to get their opinions on the strike. What is the probability that exactly of the workers interviewed are union members

Answer 1

Answer:

The probability that exaclt x of the members are union members from the n interviewed is given by:

$P(X = x) = C_{n,x}.(0.96)^{x}.(0.04)^{n-x}$

Step-by-step explanation:

For each worker, there are only two possible outcomes. Either they are union members, or they are not. Members are independent of other members. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

About 96% of all workers employed by the university belong to the workers' union.

This means that $p = 0.96$

What is the probability that exactly of the workers interviewed are union members

We want probability of x, from n workers interviewed. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = x) = C_{n,x}.(0.96)^{x}.(0.04)^{n-x}$