Probability : Quant Question Archive [LOCKED]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 16 Jan 2017, 16:55

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Probability

Author Message
Manager
Joined: 07 Nov 2004
Posts: 89
Location: London
Followers: 1

Kudos [?]: 1 [0], given: 0

### Show Tags

07 Nov 2004, 12:13
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
Posts: 4302
Followers: 40

Kudos [?]: 429 [0], given: 0

### Show Tags

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
_________________

Best Regards,

Paul

Senior Manager
Joined: 19 Oct 2004
Posts: 317
Location: Missouri, USA
Followers: 1

Kudos [?]: 78 [0], given: 0

### Show Tags

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.
_________________

Let's get it right!!!!

GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4302
Followers: 40

Kudos [?]: 429 [0], given: 0

### Show Tags

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.
_________________

Best Regards,

Paul

Manager
Joined: 07 Nov 2004
Posts: 89
Location: London
Followers: 1

Kudos [?]: 1 [0], given: 0

### Show Tags

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.
Manager
Joined: 28 Aug 2004
Posts: 205
Followers: 1

Kudos [?]: 2 [0], given: 0

### Show Tags

15 Nov 2004, 03:44
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.
15 Nov 2004, 03:44
Display posts from previous: Sort by