Given:
Probability of employees that are smokers = 35% = 0.35
Number of workers chosen random = 6
Required: Probability that there will be exactly 2 smokers
Let X denotes the number of smokers. Then X follows B(6, 0.35).
It is enough to find P(X = 2).
The binomial distribution is defined as
[tex]P(X=x)=^nC_xp^xq^{n-x}[/tex]
Substitute the given values.
[tex]P(X=x)=^6C_x(0.35)^x(0.65)^{6-x}[/tex]
To find P(X =2), plug 2 for x .
[tex]\begin{gathered} P(X=2)=^6C_2(0.35)^2(0.65)^4 \\ =\frac{6!}{2!4!}\cdot(0.0219) \end{gathered}[/tex]
