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;
![P(X = r) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r} ; n = 0,1,2,3,.......](https://tex.z-dn.net/?f=P%28X%20%3D%20r%29%20%3D%20%5Cbinom%7Bn%7D%7Br%7D%5Ctimes%20p%5E%7Br%7D%20%5Ctimes%20%281-p%29%5E%7Bn-r%7D%20%3B%20n%20%3D%200%2C1%2C2%2C3%2C.......)
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)
= ![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}](https://tex.z-dn.net/?f=1-%20%5Cbinom%7B15%7D%7B0%7D%5Ctimes%200.15%5E%7B0%7D%20%5Ctimes%20%281-0.15%29%5E%7B15-0%7D-%5Cbinom%7B15%7D%7B1%7D%5Ctimes%200.15%5E%7B1%7D%20%5Ctimes%20%281-0.15%29%5E%7B15-1%7D-%5Cbinom%7B15%7D%7B2%7D%5Ctimes%200.15%5E%7B2%7D%20%5Ctimes%20%281-0.15%29%5E%7B15-2%7D-%5Cbinom%7B15%7D%7B3%7D%5Ctimes%200.15%5E%7B3%7D%20%5Ctimes%20%281-0.15%29%5E%7B15-3%7D)
=
= <u>0.1773</u>