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October 1, 2014

October 1, 2014

Posted by **J** on Tuesday, March 6, 2007 at 5:15pm.

#1) A sample of 4 different calculators is randomly selected from a group containing 20 that are defective and 31 that have no defects. What is the probability that at least one of the calculators is defective?

(A) 0.863 (B) 0.874 (C) 0.200 (D) 0.126

My work:

(31/51)(30/50)(29/49)(28/48) = .1259

1 - .1259 = .8741

#2)A sample of 4 different calculators is randomly selected from a group containing 49 that are defective and 26 that have no defects. What is the probability that all four of the calculators selected are defective?

(A) 0.1822 (B) 0.0793 (C) 14.1724 (D) 0.1743

My work:

(49/75)(48/74)(47/73)(46/72) = .1743

#1. The probability that NONE are defective is 31/51 *30/50 * 29/49 * 28/48 = 0.1259

Probability of at least 1 = 1 - .1259 = .8741 . Correct

#2 Also correct

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