# Answer to Question #8911 in Statistics and Probability for Hasmukh

Question #8911

In the screws manufactured by a certain machine, 3 percent are found to be defective. If a

sample of 12 is taken, what is the probability that (5 Marks)

a) exactly four are defective

b) Not more than four are defective.

sample of 12 is taken, what is the probability that (5 Marks)

a) exactly four are defective

b) Not more than four are defective.

Expert's answer

In the screws manufactured by a certain machine, 3 percent are found to be defective. If a

sample of 12 is taken, what is the probability that (5 Marks)

a) exactly four are defective

The probability that the screw is defective is 0.03. Then the probability that exactly four are defective is

P = (0.03)^4·(1-0.03)^(12-4) = 6.3483·10^(-7).

b) Not more than four are defective.

P = 1 - P(0 defective) - P(1 defective) - P(2 defective) - P(4 defective) =

& & = 1 - (0.03)^0·(1-0.03)^12 -

& - (0.03)^1·(1-0.03)^(12-1) -

& - (0.03)^2·(1-0.03)^(12-2) -

& - (0.03)^3·(1-0.03)^(12-3) -

& - (0.03)^4·(1-0.03)^(12-4) = 0.284.

sample of 12 is taken, what is the probability that (5 Marks)

a) exactly four are defective

The probability that the screw is defective is 0.03. Then the probability that exactly four are defective is

P = (0.03)^4·(1-0.03)^(12-4) = 6.3483·10^(-7).

b) Not more than four are defective.

P = 1 - P(0 defective) - P(1 defective) - P(2 defective) - P(4 defective) =

& & = 1 - (0.03)^0·(1-0.03)^12 -

& - (0.03)^1·(1-0.03)^(12-1) -

& - (0.03)^2·(1-0.03)^(12-2) -

& - (0.03)^3·(1-0.03)^(12-3) -

& - (0.03)^4·(1-0.03)^(12-4) = 0.284.

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