SOLUTION: A box contains 15 items, 4 of which are defective. Three items are selected. What is the probability that the first is good, the second is defective and third is also good.

Question 1174700: A box contains 15 items, 4 of which are defective. Three items are selected. What is the probability that the first is good, the second is defective and third is also good.
Found 3 solutions by AnlytcPhil, Theo, ikleyn:
Answer by AnlytcPhil(1806)   (Show Source): You can put this solution on YOUR website!

Here is a list of all eight possible ways to select the items: first second third 1. good good good 2. good good defective 3. good defective good 4. good defective defective 5. defective good good 6. defective good defective 7. defective defective good 8. defective defective defective Notice that there was only 1 successful way, number 3, which is in red. That's 1 successful way out of 8 possible ways, so the probability is 1 out of 8, so we make that into a fraction 1/8. Answer: 1/8 Edwin

Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
there are 15 items in the box.
4 of them are defective.
11 of them are not.

the probability that the first item you pick out of the box is not defective is 11/15.

assuming the first item you picked out of the box was not defective, then the box now contains 14 items.
4 of them are defective.
10 of them are not.

the probability that the second item you pick out of the box is defective is 4/14.

assuming the second items you picked out of the box was defective, then the box now contains 13 items.
3 of them are defective.
10 of them are not.

the probability that the third item you pick out of the box is not defective is 10/13.

the probability that all 3 events occur is:

11/15 * 4/14 * 10/13 = (11*4*10) / (15*14*13) = 440 / 2730 = 4.1611721612.

Answer by ikleyn(52778)   (Show Source): You can put this solution on YOUR website!
.

In the box, there are 11 good items and 4 defective. The probability to select 1st item good is . If the 1st item was good, then the probability to get 2nd item defective is . If 1st item was good and 2nd item was defective, then the probability to get 3rd item good is . The overall probability under the problem's question is, T H E R E F O R E, P = = = = 0.1612 (rounded) = 16.12% (approximately). ANSWER

Solved.

Do not accept any other answer.

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If you ask me about the solution by respectful tutor Edwin, where and why it is wrong, I will answer you:

It is wrong, BECAUSE Edwin mistakenly consider that 8 events in his list as EQUALLY LIKELIHOOD, while they ARE NOT.

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