Question 872700: Out of each batch of 15 items, 4 are faulty. Items are examined one by one, and items checked are not replaced.
a) what is the probability that there are exactly 3 defectives in the first 8 examined
b) probability that the 9th item examined is the 4th defective one found
I can calculate probability that first item being defective is 4 in 15, and then assuming a faulty item is chosen, the second item has a probability of 3 in 14, or 4 in 14 if the first item selected works. I am sure that I could perform very many multiple calculations and maybe get the correct result, but I am sure there must be a single equation to calculate these that I have missed. I can't find it in my books, and I cannot identify the 'type' of equation to Google examples.
Any help would be very much appreciated, as my son is trying to complete an IB past paper and between us we are stuck on this one.
Gary Lynk
Found 2 solutions by stanbon, ewatrrr:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Out of each batch of 15 items, 4 are faulty. Items are examined one by one, and items checked are not replaced.
a) what is the probability that there are exactly 3 defectives in the first 8 examined
This is a Binomial Problem with n = 8 and p(defect) = 4/15
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There are 8C3 = 56 ways to have 3 of 8 defective items
The probability of eachset of 3 defect and 5 not defect i (4/15)^3*(11/15)^8
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Answer to your Problem::
P(x = 3 defect in 8) = 56(4/15)^3*(11/15)^8 = 0.2252
If you use a TI calculator you get binompdf(8,4/15,3) = 0.2252
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b) probability that the 9th item examined is the 4th defective one found
Patterm:: nnnnnnnd = (11/15)^7*(4/15) = 0.000000017168..
Cheers,
Stan H.
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Answer by ewatrrr(24785)  (Show Source):
You can put this solution on YOUR website!
Hi
Using a Binomial Distribution:
p(def) = 4/15 = .2667 and q = 1 -.2667 = .7333
a. n = 8, P(x = 3) = 56(.2667)^3(.7333)^5 0r binompdf(8, .2267, 3) = .2252
b. n = 9, P(x = 4) = 126(.2667)^4(.7333)^5 0r binompdf(9, .2267,4) = .1351
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