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) = (round to three decimal places as needed.)

Explanation:

Step1: Identify the binomial - probability formula

The binomial - probability formula is $P(x)=C(n,x)\times p^{x}\times(1 - p)^{n - x}$, where $n$ is the number of trials, $x$ is the number of successes, $p$ is the probability of success on a single trial, and $C(n,x)=\frac{n!}{x!(n - x)!}$. Here, $n = 10$, $p=0.58$, and $1 - p = 0.42$.

Step2: Calculate $C(10,5)$ for part (a)

$C(10,5)=\frac{10!}{5!(10 - 5)!}=\frac{10!}{5!5!}=\frac{10\times9\times8\times7\times6}{5\times4\times3\times2\times1}=252$.

Step3: Calculate $P(5)$ for part (a)

$P(5)=C(10,5)\times(0.58)^{5}\times(0.42)^{5}=252\times(0.58)^{5}\times(0.42)^{5}\approx252\times0.0656\times0.013069\approx0.215$.

Step4: Calculate $P(x\geq6)$ for part (b)

$P(x\geq6)=P(6)+P(7)+P(8)+P(9)+P(10)$.
$C(10,6)=\frac{10!}{6!(10 - 6)!}=210$, $P(6)=210\times(0.58)^{6}\times(0.42)^{4}\approx210\times0.0381\times0.031117\approx0.246$.
$C(10,7)=\frac{10!}{7!(10 - 7)!}=120$, $P(7)=120\times(0.58)^{7}\times(0.42)^{3}\approx120\times0.0221\times0.074088\approx0.197$.
$C(10,8)=\frac{10!}{8!(10 - 8)!}=45$, $P(8)=45\times(0.58)^{8}\times(0.42)^{2}\approx45\times0.0128\times0.1764\approx0.102$.
$C(10,9)=\frac{10!}{9!(10 - 9)!}=10$, $P(9)=10\times(0.58)^{9}\times(0.42)^{1}\approx10\times0.0074\times0.42\approx0.031$.
$C(10,10)=\frac{10!}{10!(10 - 10)!}=1$, $P(10)=(0.58)^{10}\approx0.004$.
$P(x\geq6)\approx0.246 + 0.197+0.102 + 0.031+0.004=0.580$.

Step5: Calculate $P(x\lt4)$ for part (c)

$P(x\lt4)=P(0)+P(1)+P(2)+P(3)$.
$C(10,0)=\frac{10!}{0!(10 - 0)!}=1$, $P(0)=(0.42)^{10}\approx0.00014$.
$C(10,1)=\frac{10!}{1!(10 - 1)!}=10$, $P(1)=10\times(0.58)^{1}\times(0.42)^{9}\approx10\times0.58\times0.00033=0.002$.
$C(10,2)=\frac{10!}{2!(10 - 2)!}=45$, $P(2)=45\times(0.58)^{2}\times(0.42)^{8}\approx45\times0.3364\times0.00079=0.012$.
$C(10,3)=\frac{10!}{3!(10 - 3)!}=120$, $P(3)=120\times(0.58)^{3}\times(0.42)^{7}\approx120\times0.1951\times0.00188=0.044$.
$P(x\lt4)\approx0.00014 + 0.002+0.012+0.044 = 0.058$.

Answer:

(a) $0.215$
(b) $0.580$
(c) $0.058$