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Probability

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Source: — Problem Solving |
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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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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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by earth@work » Tue Dec 02, 2008 12:05 pm
answer same as mals24 :6/11 option C
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