Answer :
Answer:
See the below works.
Step-by-step explanation:
To find the probability that all 4 selected bolts are defective of not defective, we can use the Binomial Distribution:
[tex]\boxed{P(X=x)=\left(\begin{array}{c}n\\x\end{array}\right) p^xq^{(n-x)}}[/tex]
where:
- [tex]P(X=x)=\texttt{binomial probability}[/tex]
- [tex]x=\texttt{number of success}[/tex]
- [tex]n=\texttt{number of trials}[/tex]
- [tex]p=\texttt{success rate}[/tex]
- [tex]q=\texttt{failure rate}\ \Longrightarrow\ \boxed{q=1-p}[/tex]
Let the success event (X) = bolt is defective, then the success rate (p):
[tex]\displaystyle p=\frac{4}{42}[/tex]
[tex]\displaystyle p=\frac{2}{21}[/tex]
and the failure rate (q):
[tex]q=1-p[/tex]
[tex]\displaystyle q=1-\frac{2}{21}[/tex]
[tex]\displaystyle q=\frac{19}{21}[/tex]
(a) probability that all 4 selected bolts are defective:
- [tex]n=4[/tex]
- [tex]x=4[/tex]
[tex]\begin{aligned} P(X=4)&=\left(\begin{array}{c}4\\4\end{array}\right)\left( \frac{2}{21}\right) ^4\left(\frac{19}{21} \right)^{(4-4)}\\\\&=\frac{4!}{4!(4-4)!}\left( \frac{2}{21}\right) ^4\left(\frac{19}{21} \right)^0\\\\&=1\left( \frac{2}{21}\right) ^4(1)\\\\&=\frac{16}{194481} \end{aligned}[/tex]
(b) probability that none of the 4 selected bolts are defective:
- [tex]n=4[/tex]
- [tex]x=0[/tex]
[tex]\begin{aligned} P(X=0)&=\left(\begin{array}{c}4\\0\end{array}\right)\left( \frac{2}{21}\right) ^0\left(\frac{19}{21} \right)^{(4-0)}\\\\&=\frac{4!}{0!(4-0)!}\left( \frac{2}{21}\right) ^0\left(\frac{19}{21} \right)^4\\\\&=1(1)\left( \frac{19}{21}\right) ^4\\\\&=\frac{130321}{194481} \end{aligned}[/tex]
