Probability of taking 4 defective capsules

[jzee08]

The product is a box containing 16 cold capsules. Some of the
manufactured capsules are defective (over weight, OW). A random 32
box (512 capsule) sample of the entire 1,200 boxes of capsules were
weighed. The results showed that only 1 out of every 512 capsules weigh
too much.

The instructions on the box are to take 1 capsule on day 1, 1
capsule on day 2, 2 capsules on day 3, 2 capsules on day 4, 3
capsules on day 5, 3 capsules on day 6 and the remaining 4 capsules
on day 7. What is the probability of getting 4 OW capsules on day 7?

What I have so far;
To calculate the probability of 1 OW capsule in the boxes sampled,
1/512 = 0.0019. To calculate the probability of 1 OW capsule
in one box = 1/512*16 = 0.03125. The probability of getting 4 OW
capsules in a bos would be 0.03125 * 4/16 = 0.0078 How do you look
at the use of the capsules over the 7 days?

[Thoughts]
I think the probability of getting 4 OW capsules in one box is
0.03125 * 4/16 = 0.0078. The most difficult part is how to look at
the use of the capsules over the 7 days. To find the probability of
being left with 4 OW capsules on day 7 this would be 1 of the possible outcomes from all total outcomes, on days 1, 2, 3, 4, 5, 6 and 7ie. 1/((2^1)+(2^1)+(2^2)+(2^2)+(2^3)+(2^3)+(2^4)), which is 1/44. Would the overall probability be 0.0078 * 1/44? It seems the actual probability should be much less than this. Please help

 

[SEngstrom]

First off - if 512 capsules were tested and only one was found OW that does not give a lot of confidence in the estimate of the OW rate... but that is not your problem I suspect.

It seems the problem formulation is a little bit deceptive. The most reasonable way to view the "drawing of the pills" is to say that each is an independent event with the probability p=1/512 to succeed (draw an OW pill).

Day seven is independent of the previous days since we don't know if OW pills are drawn or not during day 1-6. All that matters is the risk of drawing 4 out of 4 OW pills on day 7. That probability is simply p^4 - a pretty slim chance...

Your reasoning for 4 pills in a box is not right - if one OW pill is a rare event, then two will be even more unlikely and so on. Makes sense?

 

[jzee08]

SEngstrom

That makes sense. I seem to have been overcomplicating the problem. Thank you for your help! BTW, this is a real life question from a pharmaceutical company.

 

[SEngstrom]

If it is a real-life example it may be worth bringing up the inadequate sampling employed to determine the fault rate for those pills...

 

[CRGreathouse]

If it is a real-life example it may be worth bringing up the inadequate sampling employed to determine the fault rate for those pills...

Right. A 95% confidence Agresti-Coull interval says that the probability could be anywhere from 0% to 1.49% of finding a bad pill. You need more samples! If you had 10 out of 5120 then it would say 0.089% to 0.399%.

Of course even with only 512 pills sampled, that still gives an upper bound of 0.00000492% chance of getting four overweight pills on the last day.