# Conditional Probability of defective bulb

1. Oct 4, 2007

### Kites

I say urgent because of the horribly small lecture I received on this section, a whole 3 minutes or so of examples. While I won't give further context I can say without a doubt I am completely lost.

Here is the problem I am stuck on.

In a string of 12 Christmas tree light bulbs, 3 are defective. The bulbs are selected at random and tested, 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

I am at a lost. My only feeble attempt at a solution has been: (1/12)*(1/11)*(1/10), which I know for a fact is wrong. If someone could shed some light, any light, on how to tackle problems like this I would be eternally grateful.

2. Oct 4, 2007

### Gib Z

Sorry if its meant to be obvious, but are the 3 defective ones next to each other :( ?

3. Oct 4, 2007

### Uther

Assuming that a given bulb is not tested more than once (not said in the text, but that seems obvious to me)...

Try to word the problem in a different manner:

Case a) means that your first tested bulb turn out to be a defective one (how many chances for this to happen), same fate for the second one (again, how many chances, given there are only 2 faulty bulbs remaining) as well as for the third (chances?).

Case b) means that your fifth tested bulb turn out to be a defective one, while two out of the four previous ones were also defective. That is:

and so on

Try to express all of this in a mathematical fashion and you should see an emerging pattern.

Enjoy.

4. Oct 4, 2007

### arildno

On a), you should have 3/12*2/11*1/10.
(Three chances out of twelve that the first one is defective, then two chances out of 11 for the second to be defective as well, and lastly, one chance out of 11 to get the last defective bulb as the third one.)
Note that this is just according to the binomial formula:
$$P=\frac{\binom{3}{3}}{\binom{12}{3}}=\frac{1}{\frac{12!}{3!9!}}=\frac{3!}{12*11*10}=\frac{3}{12}*\frac{2}{11}*\frac{1}{10}$$

5. Oct 4, 2007

### Kites

I understand the above for A). Now, I must ask the question, how does this formula change to when it takes 5 attempts to find the third bulb? For b). I attempted just changing the above to:

5

3
---- Obviously this isn't correct, but since I have had little lecture on this
12 I have to ask, why? How does it change in conceptual and formula?

3

6. Oct 4, 2007

### D H

Staff Emeritus
Finding the third defective on the fifth attempt means that you have found exactly two defective bulbs in the first four attempts and then on fifth attempt you find the one remaining defective bulb. The last part (finding the one remaining defective bulb on the fifth attempt) should be easy. You could solve the first part by brute force; just enumerate all the possibilities. There is a better way ...

Last edited: Oct 4, 2007
7. Oct 4, 2007

### Kites

I see how there must be a better way, but my issue is I don't know that better way, the prof. decided to only give this section about 3 minutes of her time ( -_- ). How do I go about adjusting this? I could go with brute force solve but it wouldn't help me learn this method.

8. Oct 4, 2007

### D H

Staff Emeritus
Hmm. The enumeration technique is not so bad. Combinations with duplicates is a lot trickier than combinations of unique objects.

An alternative to explicitly enumerating all the possibilities is to use recursion. Denote {d,t} as finding exactly d defective bulbs in t attempts. Two distinct events lead to this case:
• Finding d-1 defective bulbs in t-1 attempts and an additional defective bulb on the tth attempt.
• Finding d defective bulbs in n-1 attempts and a non-defective bulb on the tth attempt.

• Now denote $p_{d,t}$ as the probability of finding exactly d defective bulbs in t attempts. The two cases outlined above become a recursive formula.

9. Oct 4, 2007

### Kites

So what you're saying DH is that it will lead to a formula that is similar to Arildno's?

If that is true then that is what I need.

10. Oct 4, 2007

### D H

Staff Emeritus
Do you know what a recursive formula is? For example, n!=n*(n-1)!

In this case, the probability of finding d defective bulbs on or before the tth attempt, $p_{d,t}$, can be defined in terms of $p_{d-1,t-1}$ and $p_{d,t-1}$. It is not a closed form solution but it is easily computable.

11. Oct 5, 2007

### Uther

Case b)

Even without knowing the proper formula, consider this:

The probability of finding the 3rd defective bulb on the 5th attempt, knowing that you have already found 2 is 1/8, because only 8 bulbs remain, and only 1 of them is defective.

Now...
The probability of finding 2 defective bulbs in the first four attempts is
etc.

The probability of the sequence B,B,G,G is (3/12)*(2/11)*(9/10)*(8/9)
The probability of the sequence B,G,B,G is (3/12)*(9/11)*(2/10)*(8/9)
etc.

All these elementary probabilities are the same: (3*2*9*8)/(12*11*10*9)
That's because 3*2 represents the decreasing number of faulty bulbs, 9*8 represents the decreasing number of good ones, and 12*11*10*9 represents the decreasing number of total untested bulbs.

Then you only need to figure out the number of ways to sort the sequence B,B,G,G which is 6.

This "6" also comes from the binomial formula, or more accurately from the number of combinations, which in this case is: 4! / (2!(4-2)!)

The total probability of b) is then:

(1/8) * (3*2*9*8)/(12*11*10*9) * 6

Last edited: Oct 5, 2007