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# Probability

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Manager
Joined: 07 Nov 2004
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Probability [#permalink]  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?
GMAT Club Legend
Joined: 15 Dec 2003
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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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Paul

Senior Manager
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Re: Probability [#permalink]  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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GMAT Club Legend
Joined: 15 Dec 2003
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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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Paul

Manager
Joined: 07 Nov 2004
Posts: 89
Location: London
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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.
Manager
Joined: 28 Aug 2004
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1 â€“ (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 â€“ 7/55 = 48/55.
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