There are 10 components, 2 are defective.
Therefore 10-2=8 components are not defective.
a) Number of ways to choose 5 non defective components is "C(8, 5)"
Number of ways to choose 5 components in total is "C(10, 5)"
Therefore
P(5 non defective components selected) = "\\frac{C(8, 5)}{C(10,2)}=\\frac{\\frac{8!}{5!3!}}{\\frac{10!}{5!5!}}=\\frac{2}{9}"
b) Exactly one defective component being included: there is one defective and 4 non defective components
Number of ways to choose 4 non defective components is "C(8, 4)"
Number of ways to choose 1 defective component is "C(2, 1)=2"
Number of ways to choose 5 components in total is "C(10, 5)"
Therefore
P(exactly one defective component being included) = "\\frac{2\\cdot C(8,4)}{C(10,5)}=\\frac{\\frac{2\\cdot8!}{4!4!}}{\\frac{10!}{5!5!}}=\\frac{5}{9}"
Answer:
a) "\\frac{2}{9}"
b) "\\frac{5}{9}"
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