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 = Number of adults who use their smartphones
The above situation can be represented through the binomial distribution;
[tex]P(X = r) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r} ; n = 0,1,2,3,.......[/tex]
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 [tex]\geq[/tex] 4)
P(X [tex]\geq[/tex] 4) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3)
= [tex]1- \binom{15}{0}\times 0.15^{0} \times (1-0.15)^{15-0}-\binom{15}{1}\times 0.15^{1} \times (1-0.15)^{15-1}-\binom{15}{2}\times 0.15^{2} \times (1-0.15)^{15-2}-\binom{15}{3}\times 0.15^{3} \times (1-0.15)^{15-3}[/tex]
= [tex]1- (1\times 1\times 0.85^{15})-(15\times 0.15^{1} \times 0.85^{14})-(105 \times 0.15^{2} \times 0.85^{13})-(455 \times 0.15^{3} \times 0.85^{12})[/tex]
= 0.1773