54​% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is ​ (a) exactly​ five, (b) at least​ six, and​ (c) less than four.

Accepted Solution

Answer:[tex]P(5)=0.238[/tex][tex]P(x\geq 6)=0.478[/tex][tex]P(x<4)=0.114[/tex]Step-by-step explanation:In this case we can calculate the probability using the binomial probability formula[tex]P(X=x)=\frac{n!}{x!(n-x)!}*p^x*(1-p)^{n-x}[/tex]Where p is the probability of obtaining a "favorable outcome " x is the number of desired "favorable outcome " and n is the number of times the experiment is repeated. In this case n = 10 and p = 0.54.(a) exactly​ fiveThis is: [tex]x=5,\ n=10,\ p=0.54.[/tex]So:[tex]P(X=5)=\frac{10!}{5!(10-5)!}*0.54^x*(1-0.54)^{10-5}[/tex][tex]P(5)=0.238[/tex](b) at least​ sixThis is: [tex]x\geq 6,\ n=10,\ p=0.54.[/tex][tex]P(x\geq 6)=P(6) + P(7)+P(8)+P(9) + P(10)[/tex][tex]P(x\geq 6)=0.478[/tex](c) less than fourThis is: [tex]x< 4,\ n=10,\ p=0.54.[/tex][tex]P(x<4)=P(3) + P(2)+P(1)+P(0)[/tex][tex]P(x<4)=0.114[/tex]