Answer:
The probability that at least 4 of them use their smartphones is 0.1773.
Step-by-step explanation:
We are given that when adults with smartphones are randomly selected 15% use them in meetings or classes.
Also, 15 adult smartphones are randomly selected.
Let X = <em>Number of adults who use their smartphones</em>
The above situation can be represented through the binomial distribution;

where, n = number of trials (samples) taken = 15 adult smartphones
r = number of success = at least 4
p = probability of success which in our question is the % of adults
who use them in meetings or classes, i.e. 15%.
So, X ~ Binom(n = 15, p = 0.15)
Now, the probability that at least 4 of them use their smartphones is given by = P(X
4)
P(X
4) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3)
= 
=
= <u>0.1773</u>