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A manufacturing plant produces a special kind of car part.

Each car part produced has a 0.087 probability of being defective.
Seven car parts are chosen at random from the production line.

To the nearest thousandth, what is the probability that exactly three of the seven-car parts will be defective?

A. 0.096
B. 0.017
C. 0.009
D. 0.002

User Angelisa
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1 Answer

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The probability that exactly x number of cars will be defective is given by


P(X)={ ^nC_xp^x(1-p)^(n-x)

Here p = 0.087, n = 7, x = 3


P(3)={ ^7C_3(0.087)^3(1-0.087)^(7-3)} \\ \\ =35(0.087)^3(0.913)^4=35(0.000658503)(0.694837) \\ \\ =0.016
User Jyr
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