Math
posted by J .
My question is to get a better understanding of this. Basically, these two problems, I did it WITHOUT replacement. So I would multiply each defects and so forth. My question is, would I have to subtract the total from 1 in the second question? I am not exactly clear on when to subtract from 1 and not to...
#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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