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A factory received a shipment of 42 bolts, and the vendor who sold the items knows there are 4 bolts in the
shipment that are defective. Before the receiving foreman accepts the delivery, he samples the shipment,
and if too many of the bolts in the sample are defective, he will refuse the shipment.
For each of the following, give your responses as reduced fractions.
If a sample of 4 bolts is-selected, find the probability that all in the sample are defective.
1
1076
If a sample of 4 bolts is selected, find the probability that none in the sample are defective.


Sagot :

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]

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