Solving Probability Problem: 5th Good Item on 9th Test

  • Thread starter eku_girl83
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In summary, the question asks for the probability of finding the fifth good item on the ninth test, given that there are 6 good and 8 defective light bulbs in a box. This can be solved by considering the first 9 bulbs as a group, with 5 good and 4 defective bulbs. The probability of this happening is calculated by using combinations and is multiplied by the probability of the ninth bulb being good, which is 5/9. The final answer is (4/6 * 6C4 * 8C4 / 14C9) * 5/9. However, the solution provided by another individual is slightly better and more consistent with negative binomial examples.
  • #1
eku_girl83
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Here's the question:
A box contains 6 good and 8 defective light bulbs. The bulbs are drawn out one at a time, without replacement, and tested. What is the probability that the fifth good item is found on the ninth test?

Could someone explain how I would go about solving this problem? Thanks!
 
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  • #2
If I told you the probability that exactly 4 good items have been found within 8 tests was 0.68, could you solve the problem?

(p.s. 0.68 is probably wrong)
 
  • #3
Consider the first 9 balls, this can be done 14C9 ( from 14 choose 9 its on your calculator). If the 9th ball is the 5th good then the first 9 balls must consist of 5 good and 4 bad balls.
The probability of this happening is 6C5*8C4/14C9. If this is true you need the 9th ball to be good. This has probably 5/9.
So the probability is 6C5*8C4/14C9 * 5/9.
 
  • #4
Not sure about that answer, Damned.

You want 4 good and 4 bad on the first 8, then to draw a bad on the 9th, which is to draw on of the 4 remaining bad ones from the 6 that are left.

[tex]\frac{\frac{4}{6}\binom{6}{4}\binom{8}{4}}{\binom{14}{8}}[/tex]but they might well be the same after simplifying
 
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  • #5
The fifth good item has to found on the 9th test. So you should replace the 4/6 with a 2/6 and this can be rearranged to give my answer. You solution is slighty better and more consistent with student examples of negative binomial etc.
 
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  • #6
Sorry for switching things over, and yes I agree with your answer entirely now I've thought about it for a second. I also agree that such conditional probabilities would be beyond the scope of the course I imagine the OP is doing.
 

1. What is the probability of getting the 5th good item on the 9th test?

The probability of getting the 5th good item on the 9th test depends on the total number of items and the number of good items in the test. If there are 10 items in total and 5 of them are good, the probability would be 1/2 or 50%. However, if the number of good items and total items differ, the probability would be different.

2. How do you solve probability problems involving finding the nth good item on the nth test?

To solve probability problems involving finding the nth good item on the nth test, you need to know the total number of items and the number of good items in the test. Then, you can use the formula: P(nth good item on nth test) = (number of good items) / (total number of items).

3. Can the probability of getting the 5th good item on the 9th test be greater than 1?

No, the probability of an event cannot be greater than 1. A probability of 1 means that the event is certain to happen, while a probability of 0 means that the event is impossible. Therefore, the probability of getting the 5th good item on the 9th test cannot be greater than 1.

4. How do you determine the number of good items and total items in a probability problem?

The number of good items and total items in a probability problem can be determined by reading the problem statement carefully. If the problem does not specify the number of good items or total items, you can assume that they are equal or use variables to represent them.

5. Can the probability of getting the 5th good item on the 9th test change if the items are replaced after each test?

Yes, the probability can change if the items are replaced after each test. This is because the probability of getting a good item on the 5th try would be affected by the previous outcomes. If the item is replaced after each test, the probability would remain the same for each try since the previous outcomes would not affect the current one.

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