Question

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.

Answer 1

Answer:

$P(5)=0.238$

$P(x\geq 6)=0.478$

$P(x$

Step-by-step explanation:

In this case we can calculate the probability using the binomial probability formula

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

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​ five

This is:

$x=5,\ n=10,\ p=0.54.$

So:

$P(X=5)=\frac{10!}{5!(10-5)!}*0.54^x*(1-0.54)^{10-5}$

$P(5)=0.238$

(b) at least​ six

This is: $x\geq 6,\ n=10,\ p=0.54.$

$P(x\geq 6)=P(6) + P(7)+P(8)+P(9) + P(10)$

$P(x\geq 6)=0.478$

(c) less than four

This is: $x< 4,\ n=10,\ p=0.54.$

$P(x$

$P(x$

Answer 2

Answer with Step-by-step explanation:

We are given that

U.S adults have very little confidence in newspaper=54%

The  probability of getting U.S adults have very little confidence in newspaper, p=0.54

The probability of getting U.S adults have not very little confidence in newspaper, q=1-0.54=0.46

Total number of  selected U.S adults=10

Binomial probability  formula

$P(X=x)=nC_x q^{n-x}p^x$

a.Substitute x=5, n=10, p=0.54, q=0.46

$P(X=5)=10C_5(0.46)^5(0.54)^5=\frac{10!}{5!5!}(0.46)^5(0.54)^5$

By using $nC_r=\frac{n!}{r!(n-r)!}$

$P(X=5)=\frac{10\times 9\times 8\times 7\times 6\times 5!}{5\times 4\times 3\times 2\times 1}{5!}(0.46)^5(0.54)^5=9\times 2\times 2\times 7(0.46)^5(0.54)^5=0.238$

The probability that number of U.S adults who have very little confidence in newspaper is exactly five=0.238

b.$P(X\geq 6)=p(X=6)+p(X=7)+p(X=8)+p(X=9)+p(X=10)$

$P(X\geq 6)=10C_6(0.46)^4(0.54)^6+10C_7(0.46)^3(0.54)^7+10C_8(0.46)^2(0.54)^8+10C_9(0.46)(0.54)^9+10C_{10}(0.54)^{10}$

$P(X\geq 6)=\frac{10\times 9\times 8\times 7\times 6!}{6!\times 4\times 3\times 2\times 1}{0.46)^4(0.54)^6+\frac{10\times 9\times 8\times 7!}{7!\times 3\times 2\times 1}(0.46)^3(0.54)^7+\frac{10\times 9\times 8!}{8!2\times 1}(0.46)^2(0.54)^8$+$\frac{10\times 9!}{9!}(0.46)(0.54)^9+\frac{10!}{10!}(0.54)^{10}$

$P(X\geq 6)=210(0.46)^4(0.54)^6+120(0.46)^3(0.54)^7+45(0.46)^2(0.54)^8+10(0.46)(0.54)^9+(0.54)^{10}$

$P(X\geq 6)=0.478$

The probability that number of U.S adults who have very little confidence in newspaper is at least six=P$P(X\geq 6)=0.478$

c.P(X<4)=p(x=0)+p(x=1)+p(x=2)=p(x=3)

$P(X$

$P(X$

$P(X$

$P(X$

Hence, the probability that number of U.S adults who have very little confidence in newspaper is less than four=P(X<4)=0.114