A shipment of computer chips contains 12 that are good and 3 that are defective. Three chips are selected in succession, with replacement and checked for a defect. find the probability that
a) the first chip will be defective, second defective , and the third good?
b) none of the chips will be defective?
c) all of the chips will be defective?
2007-05-06 15:14:03 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Because 12 out of 15 of the chips are good, the probability of drawing a good chip is 4/5, and the probability of drawing a defective chip is 3/15 = 1/5. As the selection is done with replacement, the events are independent (think of flipping a coin). This makes calculating the probabilities quite simple: we just multiply.
a.) The probability the first is defective is 1/5, same with second. The probability that the third is good is 4/5. The answer is therefore (1/5)^2 * (4/5) = 4/125.
b.) Just as before we find the probability is (4/5)^3 = 64/125.
c.) I'll leave this to you
2007-05-06 15:30:55 · answer #1 · answered by itsakitty 3 · 0⤊ 0⤋
a) (1/4)(1/4)(9/12)
b) (3/4)(3/4)(3/4)
c) (1/4)(1/4)(1/4)
2007-05-06 15:21:41 · answer #2 · answered by bruinfan 7 · 1⤊ 1⤋
a. 3/12 first is defective
2/11 defective
1/10 defective
b. none defective
9nCr3
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12nCr3
c. 3ncr3
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12nCr3
2007-05-06 15:21:26 · answer #3 · answered by leo 6 · 0⤊ 2⤋
a) 2.4%
b) 48%
c) 0.2%
2007-05-06 15:23:25 · answer #4 · answered by Baseballin11 1 · 0⤊ 1⤋