Question

Assume that when adults with smartphones are randomly selected, 61% use them in meetings or classes. If 9 adult smartphone users are randomly selectedfind the probability that exactly 5 of them use their smartphones in meetings or classes.

Answer 1

To solve this problem, let's use binominal distribution.

To calculate the probability of an event (P), use the formula below.

$P(X=x)=\frac{n!}{x!\cdot(n-x)!}\cdot p^x\cdot(1-p)^{n-x}$

Where

n = number total of events;

x = number of favorable events;

p = probability of a single event.

In this exercise:

n = 9

x = 5

p = 0.61

Then, substituting the values:

$\begin{gathered} P=\frac{9!}{5!\cdot(9-5)!}\cdot\cdot0.61^5\cdot(1-0.61)^{9-5} \\ P=\frac{9!}{5!\cdot(4)!}\cdot\cdot0.61^5\cdot(0.39)^4 \\ P=\frac{9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2}{5\cdot4\cdot3\cdot2\cdot4\cdot3\cdot2}\cdot\cdot0.61^5\cdot(0.39)^4 \\ P=\frac{9\cdot8\cdot7\cdot6}{4\cdot3\cdot2}\cdot\cdot0.61^5\cdot(0.39)^4 \\ \end{gathered}$

Solving the expression:

$\begin{gathered} P=0.2462 \\ P=24.62percent_{} \end{gathered}$

Answer: P = 0.2462.