Question

58% 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. (a) p(5)=0.216 (round to three decimal places as needed.) (b) p(x≥6)= (round to three decimal places as needed.)

Explanation:

Step1: Recall binomial probability formula

$P(x = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n = 10$, $p=0.58$, $1-p = 0.42$, and $C(n,k)=\frac{n!}{k!(n - k)!}$

Step2: Calculate $P(x\geq6)$

$P(x\geq6)=P(x = 6)+P(x = 7)+P(x = 8)+P(x = 9)+P(x = 10)$
$P(x = 6)=C(10,6)\times(0.58)^{6}\times(0.42)^{4}=\frac{10!}{6!(10 - 6)!}\times(0.58)^{6}\times(0.42)^{4}=210\times0.0388\times0.0311\approx0.252$
$P(x = 7)=C(10,7)\times(0.58)^{7}\times(0.42)^{3}=\frac{10!}{7!(10 - 7)!}\times(0.58)^{7}\times(0.42)^{3}=120\times0.0225\times0.0741\approx0.201$
$P(x = 8)=C(10,8)\times(0.58)^{8}\times(0.42)^{2}=\frac{10!}{8!(10 - 8)!}\times(0.58)^{8}\times(0.42)^{2}=45\times0.0131\times0.1764\approx0.104$
$P(x = 9)=C(10,9)\times(0.58)^{9}\times(0.42)^{1}=\frac{10!}{9!(10 - 9)!}\times(0.58)^{9}\times(0.42)^{1}=10\times0.0076\times0.42\approx0.032$
$P(x = 10)=C(10,10)\times(0.58)^{10}\times(0.42)^{0}=1\times0.0045\times1 = 0.0045$
$P(x\geq6)=0.252+0.201+0.104+0.032+0.0045 = 0.638$

Answer:

(b) $P(x\geq6)=0.638$