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Probability

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Probability [#permalink] New post 07 Nov 2004, 12:13
Interesting problem so though I'd share it with you guys.

Three items are removed from a box of 12 items. All three items are inspected. The box is rejected if more than one item is found to be faulty.
If there are three faulty items in the box, what is the probability that the box is accepted?
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 [#permalink] New post 07 Nov 2004, 12:38
Is it 39/55?
What you want is find 1 minus "rejection" which is the prob of having 0 or 1 defective in first draw. This way, of the remaining items, at least 1 will be defective and it will result in the box being rejected.
P(drawing 0 defective): 3C0*9C3 / 12C3 = 7/55
P(drawing 1 defective): 3C1*9C2 / 12C3 = 9/55
1 - (7/55 + 9/55) = 39/55
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Re: Probability [#permalink] New post 07 Nov 2004, 12:56
oxon wrote:
Interesting problem so though I'd share it with you guys.

Three items are removed from a box of 12 items. All three items are inspected. The box is rejected if more than one item is found to be faulty.
If there are three faulty items in the box, what is the probability that the box is accepted?


Let A be a bad ball and B be a good one.
Box is rejected if the 2 or 3 balls that are drawn are bad.
Possible combinations for the box to be rejected=
ABA, BAA, AAA.

Hence, the probability for the box to be rejected
(3*9*2)/(12*11*10)+ (9*3*2)/(12*11*10)+(3*2*1)/(12*11*10)=19/220

hence the probability that it will be accepted= 1-19/220=201/220.
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 [#permalink] New post 07 Nov 2004, 13:16
Oh well, I just noticed that my interpretation was that if 3 balls are removed, what is the prob that we will not have more than 1 faulty item out of the 9 remaining items in the box.
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 [#permalink] New post 07 Nov 2004, 14:01
This's how I did it.
C(12,3) = 220 combos
9 items are not faulty so # of ways to choose 3 items that are not faulty is C(9,3) = 84.
The number of possibilities of choosing one faulty item and 2 good items is C(3,1)*C(9,2) = 72.
So probability that the # of faulty items is 0 or 1 is: P = (84+72)/220 = 39/55.

Paul's note shows a better and shorter method that yields the same result.
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 [#permalink] New post 15 Nov 2004, 03:44
1 – (3C2 * 9C1 + 3C3)/12C3 = 48/55; or,

BGB + GBB + BBB + BBG (B for bad, G for Good)

(3/12*9/11*2/10) + (9/12*3/11*2/10) + (3/12*2/11*1/10) + (3/12*2/11*9/10) = 7/55

1 – 7/55 = 48/55.
  [#permalink] 15 Nov 2004, 03:44
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