This is a binomial probability distribution question.
The binomial probability formula is >>>
[tex]P(x)=\frac{n!}{(n-x)!x!}p^xq^{n-x}_{}[/tex]Where
n is the number of trials (number being sampled)
x is the number of success desired
p is the probability of getting a success
q = 1 - p is the probability of failure
From the given problem,
n = 6
x = 3
p = 58% = 0.58
q = 1 - p = 1 - 0.58 = 0.42
Substituting into the formula, we find the probability that exactly 3 of the people use smartphone in meetings or classes:
[tex]\begin{gathered} P(x)=\frac{n!}{(n-x)!x!}p^xq^{n-x}_{} \\ P(x=3)=\frac{6!}{(6-3)!3!}(0.58)^3(0.42)^{6-3}_{} \\ =\frac{6!}{3!3!}(0.58)^3(0.42)^3 \\ =0.2891 \end{gathered}[/tex]