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 4 workers (chosen at random) at the university to get their opinions on the strike. What is the probability that exactly 3 of the workers interviewed are union members?

Answer 1

Answer: 0.14155776

Step-by-step explanation:

Given : The proportion of workers employed by the university belong to the workers' union=0.96

Let x be a binomial variable that represents that worker belongs to the workers' union.

Sample size : n= 4

$P(x=3)=^4C_3(0.96)^3(0.04)^1\\\\=(4)(0.96)^3(0.04)=0.14155776$

[Binomial probability formula : P(X=x)=^nC_x(p)^x(1-p)^{n-x}]

Hence, the probability that exactly 3 of the workers interviewed are union members $=0.14155776$