Question

Assume that when with smartphones are randomly selected, 46% use them in meetings or classes. If 12 adult smartphone users are randomly selected, find the probability that fewer than 3 of them use their smartphones in meetings or classes.

Answer 1

Answer:

The probability that fewer than 3 of them use their smartphones in meetings or classes is 0.03634

Step-by-step explanation:

Apply binomial probability formula which is written as;

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

where P=46% =0.46, n=12

For this case where you find the probability that fewer than 3 of them use their smartphones in meeting or classes, x=2,1,0

Evaluating binomial probability at x=2 will be;

$P(X=2)=\frac{12!}{2!(12-2)!} *0.46^2*(1-0.46)^{12-2} \\\\P(X=2)=\frac{12!}{2!10!} *0.2116*0.002108\\\\=66*0.2116*0.002108\\=0.02944$

P(X=2)=0.02944

Evaluate binomial probability at x=1

$P(X=1)=\frac{12!}{1!(12-1)!} *0.46^1*(1-0.46)^{12-1} \\\\P(X=1)=\frac{12!}{1!11!} *0.46*0.54^{11} \\\\P(X=1)=12*0.46*0.001138\\\\P(X=1)=0.006284$

P(X=1)= 0.006284

Evaluate binomial probability at x=0

$P(X=0)=\frac{12!}{0!(12-0)!} *0.46^0*(1-0.46)^{12-0} \\\\P(X=0)=\frac{12!}{1*12!} *0.46^0*0.54^{12} \\\\P(X=0)=1*1*0.0006148 =0.0006148$

P(X=0)=0.0006148

Now for P(X<3)=P(X=2)+P(X=1)+P(X=10) =0.02944+0.006284+0.0006148

=0.0363388

=0.03634 (in 4 significant figures)