Probability

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bacali: Master | Next Rank: 500 Posts; Posts: 245; Joined: Thu Aug 16, 2007 12:41 pm

by bacali » Tue Dec 02, 2008 11:25 am

I keep getting a different answer from the OA, be advised OA might not be correct.

OA: [spoiler] C? [/spoiler]

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Source: Beat The GMAT — Problem Solving | Tue Dec 02, 2008 11:25 am

mals24: Legendary Member; Posts: 683; Joined: Tue Jul 22, 2008 1:58 pmFollowed by: 2 members; Location: Dubai; Thanked: 73 times

by mals24 » Tue Dec 02, 2008 11:55 am

Defective pens = 3
Non defective pens = 9

Probability of picking first defective pen = 9/12
Probability of picking 2nd defective pen = 8/11

Total probability = (9*8)/(12*11) = 6/11 option C

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muzali: Senior | Next Rank: 100 Posts; Posts: 95; Joined: Fri Sep 28, 2007 10:16 am; Thanked: 6 times; GMAT Score:710

by muzali » Tue Dec 02, 2008 11:56 am

Total ways of picking 2 pens that contain defective (D) or normal (N) pens are: DN, ND, NN, DD (this is just like a coin toss question)

Required Probability = 1 - (combined Prob of DN, ND and DD)
Prob of DN = (3/12)*(9/11)
Prob of ND = (9/12)*(3/11)
Prob of DD = (3/12)*(2/11)

Combined prob = (27+27+6)/(12*11) = 60/(12*11) = 5/11

Therefore, required prob = 1-(5/11) = 6/11 answer

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earth@work: Master | Next Rank: 500 Posts; Posts: 248; Joined: Mon Aug 11, 2008 9:51 am; Thanked: 13 times

by earth@work » Tue Dec 02, 2008 12:05 pm

answer same as mals24 :6/11 option C

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