Given:
A factory received a shipment of 26 sprockets, and the vendor who sold the items knows there are 5 sprockets in the shipment that are defective. Before the receiving foreman accepts the delivery, he samples the shipment, and if too many of the sprockets in the sample are defective, he will refuse the shipment.
Required:
(1) If a sample of 5 sprockets is selected, find the probability that all in the sample are defective.
(2) If a sample of 5 sprockets is selected, find the probability that none in the sample are defective.
Step-by-step explanation:
The total number of sprockets= 26
The number of defect sprockets = 5
The number of non defect sprockets = 26 - 5 = 21
(1) If a sample of 5 sprockets is selected, then the probability that all in the sample are defective is:
![\begin{gathered} P(def.)=(^5c_5)/(^(26)c_5) \\ P(def.)=((5!)/((5-5)!(5)!))/((26!)/((26-5)!(5)!)) \\ P(def.)=(1)/((26*25*24*23*22)/(5*4*3*2*1)) \\ P(def.)=(1)/(65,780) \end{gathered}]()
(2) If a sample of 5 sprockets is selected, then the probability that none in the sample are defective is:
![\begin{gathered} P(non\text{ def.\rparen=}(^(21)C_5)/(_(^26C_5)) \\ P(non\text{ def.\rparen=}(((21)!)/((21-5)!(5)!))/(((26)!)/((26-5)!(5)!)) \\ P(non\text{ def.\rparen=}((21*20*19*18*17)/(5*4*3*2*1))/((26*25*24*23*22)/(5*4*3*2*1)) \end{gathered}]()
![\begin{gathered} P(non\text{ def.\rparen=}(21*20*19*18*17)/(26*25*24*23*22) \\ P(non\text{ def.\rparen=}(20349)/(65780) \end{gathered}]()
Final Answer:
(1)
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