A tray of electronic components contains 15 components, 4 of which are defective. If 4 components are selected, what is the possibility that (a) all 4 are defective? (b) 3 are defective and 1 is good? (c) exactly 2 are defective? (d) none are defective?

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Question:

A tray of electronic components contains 15 components, 4 of which are defective. If 4 components are selected, what is the possibility that (a) all 4 are defective? (b) 3 are defective and 1 is good? (c) exactly 2 are defective? (d) none are defective?

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Answer:a) 0.0007326b) 0.03223c) 0.2418d) 0.2418Explanation:To find different probabilities for the selection of components among eleven good and four defective components, we will use the Combination.a) [tex]C(4,4) = 1; C(15,4) = 1365[/tex][tex]P = \frac{C(4,4)}{C(15,4)} = \frac{1}{1365} = 0.0007326[/tex]b) [tex]C(4,3) = 4; C(11,1) = 11[/tex][tex]P = \frac{C(4,3)*C(11,1)}{C(15,4)} = \frac{4*11}{1365} = 0.03223[/tex]c) [tex]C(4,2) = 6; C(11,2) = 55[/tex][tex]P = \frac{C(4,2)*C(11,2)}{C(15,4)} = \frac{6*55}{1365} = 0.2418[/tex]d) [tex]C(11,4) = 330[/tex][tex]P = \frac{C(11,4)}{C(15,4)} = \frac{330}{1365} = 0.2418[/tex]

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