SOLUTION: A box contains 24 widgets, 4 of which are defective. If 4 are sold at random, find the probability that a) all are defective b) none are defective.
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-> SOLUTION: A box contains 24 widgets, 4 of which are defective. If 4 are sold at random, find the probability that a) all are defective b) none are defective.
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Question 320334: A box contains 24 widgets, 4 of which are defective. If 4 are sold at random, find the probability that a) all are defective b) none are defective. Found 2 solutions by nyc_function, jrfrunner:Answer by nyc_function(2741) (Show Source):
You can put this solution on YOUR website! Part (a):
p(all are defective) = 4/24 = 1/6
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Part (b):
p(none of defective) = 1 - 4/24
p(none of defective) = 24/24 - 4/24
p(none of defective) = 20/24 = 5/6
You can put this solution on YOUR website! a) all are defective.
This means selecting four defectives in a row. The first one being defective is 4/24, the second being defective is 3/23 the third one being defective is 2/22 and the fourth one being defective is 1/21, therefore P(all defective) = P(1st defective)*P(2nd defective)*P(3rd defectie)*P(4th defective) = 4/24*3/23*2/22*1/21=24/255024 = 0.000094
b)one being defective
This means 1 defective and 3 non-defective =4C1* P(D)*P(N)*P(N)*P(N)*P(N) since there are 4C1 ways to arrange that "one" defective. 4*4/24*20/23*19/22*18/21=109440/255024=0.429
