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

Answer 1

Answer: 0.1699

Step-by-step explanation:

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

Sample size : n= 5

Binomial probability distribution formula :

$P(X=x)= \ ^nC_xp^x(1-p)^x$

The probability that exactly 4 of the workers interviewed are union members :-

$P(X=4)=^5C_4(0.96)^4(1-0.96)^1\\\\=(5)(0.96)^4(0.04)\\\\=(5)(0.84934656)(0.04)\\\\=0.169869312\approx0.1699$

Required probability = 0.1699