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 :
[tex]P(X=x)= \ ^nC_xp^x(1-p)^x[/tex]
The probability that exactly 4 of the workers interviewed are union members :-
[tex]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[/tex]
Required probability = 0.1699
