43% of men consider themselves professional baseball fans. you randomly select 10 men and ask each if he considers himself a professional baseball fan. find the probability that the number who consider themselves baseball fans is (a) exactly five, (b) at least six, and (c) less than four.
(a) ( p(5)=0.223 ) (round to three decimal places as needed.)
(b) ( p(x geq 6)=square ) (round to three decimal places as needed.)

Question

Accepted Answer

# Explanation:
## Step1: Recall the binomial probability formula
The binomial probability formula is $P(x)={n\choose x}p^{x}(1 - p)^{n - x}$, where $n = 10$ (number of trials), $p=0.43$ (probability of success), and $x$ is the number of successes.

## Step2: Calculate $P(x\geq6)$
$P(x\geq6)=P(6)+P(7)+P(8)+P(9)+P(10)$

For $x = 6$:
$$
\begin{align*}
{n\choose x}&=\frac{n!}{x!(n - x)!}=\frac{10!}{6!(10 - 6)!}=\frac{10\times9\times8\times7}{4\times3\times2\times1}=210\
P(6)&=210\times(0.43)^{6}\times(1 - 0.43)^{10 - 6}\
&=210\times(0.43)^{6}\times(0.57)^{4}\
\end{align*}
$$
$$
\begin{align*}
(0.43)^{6}&=0.43\times0.43\times0.43\times0.43\times0.43\times0.43\approx0.00632\
(0.57)^{4}&=0.57\times0.57\times0.57\times0.57\approx0.1056\
P(6)&=210\times0.00632\times0.1056\approx0.139
\end{align*}
$$

For $x = 7$:
$$
\begin{align*}
{n\choose x}&=\frac{10!}{7!(10 - 7)!}=\frac{10\times9\times8}{3\times2\times1}=120\
P(7)&=120\times(0.43)^{7}\times(0.57)^{3}\
(0.43)^{7}&=(0.43)^{6}\times0.43\approx0.00272\
(0.57)^{3}&=0.57\times0.57\times0.57\approx0.1852\
P(7)&=120\times0.00272\times0.1852\approx0.060
\end{align*}
$$

For $x = 8$:
$$
\begin{align*}
{n\choose x}&=\frac{10!}{8!(10 - 8)!}=\frac{10\times9}{2\times1}=45\
P(8)&=45\times(0.43)^{8}\times(0.57)^{2}\
(0.43)^{8}&=(0.43)^{7}\times0.43\approx0.00117\
(0.57)^{2}&=0.57\times0.57 = 0.3249\
P(8)&=45\times0.00117\times0.3249\approx0.017
\end{align*}
$$

For $x = 9$:
$$
\begin{align*}
{n\choose x}&=\frac{10!}{9!(10 - 9)!}=10\
P(9)&=10\times(0.43)^{9}\times(0.57)^{1}\
(0.43)^{9}&=(0.43)^{8}\times0.43\approx0.0005\
P(9)&=10\times0.0005\times0.57\approx0.003
\end{align*}
$$

For $x = 10$:
$$
\begin{align*}
{n\choose x}&=\frac{10!}{10!(10 - 10)!}=1\
P(10)&=(0.43)^{10}\approx0.0001
\end{align*}
$$

$P(x\geq6)=0.139 + 0.060+0.017+0.003+0.0001\approx0.219$

# Answer:
$P(x\geq6)\approx0.219$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the binomial probability formula

Step2: Calculate \(P(x\geq6)\)

Answer: