Question

Assume that when adults with smartphones are randomly selected, 58% use them in meetings or classes. If 7 adult smartphone users are randomly selected, find the probability that at least 2 of them use their
smartphones in meetings or classes.
The probability is
(Round to four decimal places as needed.),

Answer 1

Answer:

$P(x\geq 2)=0.9753$

Step-by-step explanation:

The number of adults that use smartphones in meetings or classes follows a binomial distribution. So, the probability that x adults use their  smartphones in meetings or classes is calculated as:

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

Where, n is the number of identical and independents events, in this case, 7 adults with smartphones. p is the probability of success or the probability that the adult use the smartphone in meeting or classes, in this case, p is equal to 0.58

So, replacing values, we get that the probability that x adults use their  smartphones in meetings or classes is:

$P(x)=\frac{7!}{x!(7-x)!}*0.58^x*(1-0.58)^{7-x}$

Now, the probability that at least 2 of them use their  smartphones in meetings or classes is calculated as:

$P(x\geq 2)=P(2)+P(3)+P(4)+P(5)+P(6)+P(7)$

Where:

$P(2)=\frac{7!}{2!(7-2)!}*0.58^2*(1-0.58)^{7-2}=0.0923\\P(3)=\frac{7!}{3!(7-3)!}*0.58^3*(1-0.58)^{7-3}=0.2125\\P(4)=\frac{7!}{4!(7-4)!}*0.58^4*(1-0.58)^{7-4}=0.2934\\P(5)=\frac{7!}{5!(7-5)!}*0.58^5*(1-0.58)^{7-5}=0.2431\\P(6)=\frac{7!}{6!(7-6)!}*0.58^6*(1-0.58)^{7-6}=0.1119\\P(7)=\frac{7!}{7!(7-7)!}*0.58^7*(1-0.58)^{7-7}=0.0221$

Finally, $P(x\geq 2)$ is equal to:

$P(x\geq 2)=0.0923+0.2125+0.2934+0.2431+0.1119+0.0221\\P(x\geq 2)=0.9753$