Answer:
a) The probability that a box containing 3 defectives will be shipped is [tex]51.74\%[/tex]
b) The probability that a box containing only 1 defective will be sent back for screening is [tex]17.39\%[/tex]
Step-by-step explanation:
Hi
a) The first step is to count the number of total possible random sets of taking a sample size of 4 items over 23 items of the box, so [tex]\left[\begin{array}{ccc}23\\4\end{array}\right] =23C4=8855[/tex]
The second step is to count the number of total possible random sets of taking a sample size of 4 items over 20 items of the box (discounting the 3 defectives) as the possible ways to succeed, so [tex]\left[\begin{array}{ccc}20\\4\end{array}\right] =20C4=4845[/tex]
Finally we need to compute [tex]\frac{\# ways\ to\ succeed}{\# random\ sets\ of \ 4} =\frac{4845}{8855}=0.5471=P(S)[/tex], therefore the probability that a box containing 3 defectives will be shipped is [tex]P(S)=54.71\%[/tex]
a) The first step is to count the number of total possible random sets of taking a sample size of 4 items over 23 items of the box, so [tex]\left[\begin{array}{ccc}23\\4\end{array}\right] =23C4=8855[/tex]
The second step is to count the number of total possible random sets of taking a sample size of 4 items over 22 items of the box (discounting the defective 1) as the possible ways to succeed, so [tex]\left[\begin{array}{ccc}22\\4\end{array}\right] =22C4=7315[/tex]
Then we need to compute [tex]\frac{\# ways\ to\ succeed}{\# random\ sets\ of \ 4} =\frac{7315}{8855}=0.8260=P(S)[/tex], therefore the probability that a box containing 1 defective will be shipped is [tex]P(S)=82.60\%[/tex]
Finally the probability that a box containing only 1 defective will be sent back for screening will be [tex]P(BS)=1-P(S)=1-0.8260=0.1739=17.39\%[/tex]
