SOLUTION: it is knw that there are two deffective items in a group of ten. One item is selected and then a secon without replacing the first item what is the probability that both item are g

Question 698099: it is knw that there are two deffective items in a group of ten. One item is selected and then a secon without replacing the first item what is the probability that both item are good (2)the second one is deffective (3) both are deffective
Answer by Positive_EV(69) (Show Source): You can put this solution on YOUR website!
If there are two defective items in a group of ten, there are eight good items.

1) The probability of drawing two good items is the probability the first item is good times the probability the second item is good. For the first draw, there are 8 good items out of 10 items, so the probability is 8/10 = 4/5. Since this item is removed from the box, there are now only 7 good items left, and 9 total items. The probability the second item is also good is (7/9). Thus, the probability that both items are good is (4/5)*(7/9) = 28/45.

2) Since the items are drawn without replacement, the probability the second item drawn is defective depends on whether the first item drawn is defective or not. If a working item is drawn, there will be 2 defective items out of 9 remaining, so the probability is 2/9 if the first item drawn works. If the first item drawn is defective, then there is only 1 defective item left in the box with 9 items, so the probability if the first item is defective is only 1/9 in this case.

The overall probability is the sum of the probability that the second item is defective given the first item is good and the probability that the second item is defective given the first item is also defective.

The first item is good with probability 8/10 = 4/5, and the probability that the second item is defective when the first is good is 2/9.
The second item is defective with probability 2/10 = 1/5, and the the probability that the second item is defective when the first is defective is 1/9.

The overall probability is the sum of the products of these two situations, which is (4/5)*(2/9) + (1/5)*(1/9) = 8/45 + 1/45 = 9/45 = 1/5.

3) The probability that the first and second items are both defective is equal to the product of the probability the first item is defective times the probability the second item is defective. This, as calculated in part 2), is equal to (1/5)*(1/9) = 1/45.
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