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Mathematics 14 Online
OpenStudy (anonymous):

help: A box contains 29 widgets, 4 of which are defective. If 4 are sold at random, find the probability that (a) all are defective (b) none are defective

OpenStudy (anonymous):

(4/24)*(3/23)*(2/22)*(1/21) ~.00009 ~.009% (b) (20/24)*(19/23)*(18/22)*(17/21) .45 45% Was that helpful to you?

OpenStudy (anonymous):

Or I could say it more detailed (a) If 4 are sold, probability that all the 4 are defective = (1/6)(1/6)(1/6)(1/6) = 1/(6^4) = 1/1296 = 0.00077 (b) If 4 are sold, probability that none of the 4 are defective = (5/6)(5/6)(5/6)(5/6) = (5/6)^4 = 0.48225 Now you have answer and understand?

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