Conditional Probability?

Question

In a string of 12 christmas lights, 3 are defective. Bulbs are selected at random, one at a time, until the third defective bulb is found. Compute the probability that the third defective bulb is the

A)Third bulb tested
B)Fifth bulb tested
C)Tenth bulb tested

Please explain your answers.

Thanks...

Answer 1

Edit: The poster above me has made a mistake. He has calculated the probabilities that the third defective bulb will be _one of_ the first three/five/ten bulbs tested, instead of the probability that it will be exactly the third/fifth/tenth bulb tested. Since he is calculating a different set of probabilities than those asked for by the problem, his answer is wrong.

Consider labeling the bulbs 1-12 in the order that they will be tested, and ask "of all the 12! possible labelings, which are consistent with the indicated outcome?"

A) here, only those labelings that assign the defective bulbs some permutation of the 3 numbers 1, 2, and 3, and the good bulbs some permutation of the 9 remaining numbers, are consistent with this. There are 3! ways to assign the numbers 1, 2, and 3 to the defective bulbs, and 9! ways to assign the remaining numbers to the good bulbs, so there are 3!*9! outcomes in this event out of a possible 12!. Therefore, the probability is 3!*9!/12! = 6/(12*11*10) = 1/220.

B) For this to happen, first the number 5 must be assigned to one of the defective bulbs (3 ways to do this), then two of the remaining numbers 1-4 must be assigned to the other two defective bulbs (4 possibilities for the first bulb times 3 for second, total of 12 ways to do this), and finally the remaining 9 numbers, whatever they are, must be assigned to the good bulbs (9! ways to do this). Therefore, the total probability is 3*12*9!/12! = 36/(12*11*10) = 3/110

C) For this to happen, first choose a defective bulb to receive the 10 (3 ways to do this), then for the other two defective bulbs, assign two of the numbers 1-9 to them (9*8 or 72 ways to do this), and then assign the 9 remaining numbers to the good bulbs (9! ways to do this). Then the probability is:

3*72*9!/12! = 3*72/(12*11*10) = 9/55

Answer 2

========== EDIT =====

Pascal is right, I misinterpreted the question (and I now see where I erred), in assuming that it meant finding the three defective bulbs "within" 3, 5, and 10 bulb checks respectively.

Anyway I won't change my answer, even though it's the answer to another problem. Just see Pascal's answer for the real answer to this problem though.

Nice catch, Pascal, thumbs up for that!

========== END EDIT ==========

My method will not use conditional probability.

I will use the notation nCr to mean "n choose r", this represents the way n "things" can be chosen from "r" things.

nCr = n!/[ r! * (n-r)! ], but most calculators can compute it.

Sometimes you will se the notation of two numbers in parenthesis one on top of the other (this is also known as a binomial coefficient). The n is the top number, the r is the bottom number.

========== A ==========

If we pick three bulbs there is only 1 way they are all defective ones (we assume the order they are picked in doesn't matter, as it won't for the problem). This is because (12 - 3)C(3 - 3) = 9C0 = 1. We use 12 - 3 and 3 - 3 because we are picking 3 bulbs from 12, but we're fixing 3 of them (since we're forcing these to be defective), so we're really not making a choice at all (so we're picking 0 bulbs from 9), so there's only once choice.

There are a total of 12C3 = 220 of picking 3 bulbs from 12.

So P = (successful outcomes/(all possible outcomes) = 1/220 ≈ 0.0045.

========== B ==========

Ways of picking all defective 3 bulbs (and 2 others) when choosing 5 from 12 (successful outcomes):

(12 - 3)C(5 - 3) = 9C2 = 36

We use 12 - 3 and 5 - 3 because we're only picking 2 bulbs out of 9 as we're fixing 3 of them to be the defective ones.

Ways of picking 5 bulbs out of 12 (total outcomes):

12C5 = 792

P = 36/792 = 1/22 ≈ 0.0455.

========== C ==========

Ways of picking all defective 3 defective bulbs (and 7 others) when choosing 10 from 12 (successful outcomes):

(12 - 3)C(10 - 3) = 9C7 = 36

Ways of picking 5 bulbs out of 12 (total outcomes):

12C10 = 66

P = 36/66 = 6/11 ≈ 0.5455.

Answer 3

c. In order for the 10th bulb to be the 3rd defective bulb, then out of the first nine selected, there must be 2 defective bulbs and 7 working bulbs selected. Then, the 10th must be the one defective bulb of the remaining 3.

There are (3c2)(9c7) ways to pick 2 defective and 7 working bulbs from the sample. There are (12c9) total ways to pick any 9 bulbs (working or not) from the sample of 12. The odds of getting exactly 2 defective bulbs after selecting 9 total is thus:
(3c2)(9c7)/(12c9) = 3x36/220 = 108/220.

Finally, after selecting 9, we must select the one remaining defective bulb out of the remaining 3 total. The odds of this is 1/3.

Total odds = (108/220)(1/3) = 16.4%

Hopefully, A and B are easy from this point. You should get 1/220 for A and 27/990 for B.


Note: (XcY) = X!/Y!(X-Y)!