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tough lot, please bear

by sanju09 » Mon Feb 22, 2010 2:05 am
A lot contains 20 articles. The probability that the lot contains exactly 2 defective articles is 0.4; and the probability that it contains exactly 3 defective articles is 0.6. Articles are drawn from the lot at random, one by one, without replacement until all the defective articles are found. What is the probability that the testing procedure stops at the twelfth testing?

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by firdaus117 » Mon Feb 22, 2010 2:23 am
Hey Sanju,this is a case of conditional probability and here I think Bayes Theorem should be used.It is a lenghty question but let me have a try at it.
Case 1
When lot contains two defective items,we have
P(D/L2)=Probability of getting second defective article in twelfth attempt
Also for scanning for defective article to stop at twelfth go will mean that one/two article(s) has been found in first 11 tries,

P(D/L2)=(2C1*18C10/20C11)*1/9
Similarly for case 2

P(D/L3)=(3C2*17C9/20C11)*1/9
Since two cases are mutually exclusive,
Total probability=0.4*P(D/L2)+0.6P(D/L3) (Note P(L2)=0.4 and P(L3)=0.6 )
Solving that I am getting 99/1900.
Can we have the options please?????
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by sanju09 » Mon Feb 22, 2010 2:55 am
firdaus117 wrote:Hey Sanju,this is a case of conditional probability and here I think Bayes Theorem should be used.It is a lenghty question but let me have a try at it.
Case 1
When lot contains two defective items,we have
P(D/L2)=Probability of getting second defective article in twelfth attempt
Also for scanning for defective article to stop at twelfth go will mean that one/two article(s) has been found in first 11 tries,

P(D/L2)=(2C1*18C10/20C11)*1/9
Similarly for case 2

P(D/L3)=(3C2*17C9/20C11)*1/9
Since two cases are mutually exclusive,
Total probability=0.4*P(D/L2)+0.6P(D/L3) (Note P(L2)=0.4 and P(L3)=0.6 )
Solving that I am getting 99/1900.
Can we have the options please?????
I am checking if the thinkers can deduce something like that theorem on their own, without knowing or naming Mr Bayes, if impossible, kindly ignore the sum for now.
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by firdaus117 » Mon Feb 22, 2010 3:24 am
sanju09 wrote: I am checking if the thinkers can deduce something like that theorem on their own, without knowing or naming Mr Bayes, if impossible, kindly ignore the sum for now.
Dude,the questions on probability are generally encountered by the people who score 49-51 in GMAT Maths.If you are not aiming that then you can safely let Mr Bayes rest in peace in his tomb and also ignore topics like conditional probability and reverse probability.
:)
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by ajith » Mon Feb 22, 2010 3:34 am
sanju09 wrote:A lot contains 20 articles. The probability that the lot contains exactly 2 defective articles is 0.4; and the probability that it contains exactly 3 defective articles is 0.6. Articles are drawn from the lot at random, one by one, without replacement until all the defective articles are found. What is the probability that the testing procedure stops at the twelfth testing?

Choices are under construction, tough lot, please bear.
The lot either contains 2 defectives or 3 defectives with probability 0.4 and 0.6 respectively

Say the lot contains 3 defectives, the probability that all of the three would be selected in first 12 = 17C9*3C3/20C12 =0.192

Say the lot contains 2 defectives, the probability that all of the three would be selected in first 12 = 18C10*2C2/20C12 =0.347


probability that the testing procedure stops at the twelfth testing = 0.6*.192 +0.4*.347 = 0.254
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by sanju09 » Mon Feb 22, 2010 3:35 am
firdaus117 wrote:
sanju09 wrote: I am checking if the thinkers can deduce something like that theorem on their own, without knowing or naming Mr Bayes, if impossible, kindly ignore the sum for now.
Dude,the questions on probability are generally encountered by the people who score 49-51 in GMAT Maths.If you are not aiming that then you can safely let Mr Bayes rest in peace in his tomb and also ignore topics like conditional probability and reverse probability.
:)
TRUE, but we cannot openly suggest the common GMAT aspirants to know or use the terms and names that are not introduced to them until their 10th standard mathematics. We can however make them think (may be) more than the NOT introduced concepts using their common wits that they could have had until their 10th standard mathematics.
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by firdaus117 » Mon Feb 22, 2010 3:41 am
ajith wrote:
sanju09 wrote:A lot contains 20 articles. The probability that the lot contains exactly 2 defective articles is 0.4; and the probability that it contains exactly 3 defective articles is 0.6. Articles are drawn from the lot at random, one by one, without replacement until all the defective articles are found. What is the probability that the testing procedure stops at the twelfth testing?

Choices are under construction, tough lot, please bear.
The lot either contains 2 defectives or 3 defectives with probability 0.4 and 0.6 respectively

Say the lot contains 3 defectives, the probability that all of the three would be selected in first 12 = 17C9*3C3/20C12 =0.192

Say the lot contains 2 defectives, the probability that all of the three would be selected in first 12 = 18C10*2C2/20C12 =0.347


probability that the testing procedure stops at the twelfth testing = 0.6*.192 +0.4*.347 = 0.254
Hey Ajith,
I beg to differ from your solution.In your solution,you have calculated probability of getting 2 defective articles in 12 attempts,it may be in any two from 1 to 12th attempt which is not true.But as soon as 2nd defective item is to be found it has to be stopped here which accd to question is in the twelfth attempt.Similarly for 3 defective items......
Rather you should find 1 defective article in 11 attempts and in twelfth the second defective item.That is how I have done it.
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by firdaus117 » Mon Feb 22, 2010 3:48 am
sanju09 wrote:
firdaus117 wrote:
sanju09 wrote: I am checking if the thinkers can deduce something like that theorem on their own, without knowing or naming Mr Bayes, if impossible, kindly ignore the sum for now.
Dude,the questions on probability are generally encountered by the people who score 49-51 in GMAT Maths.If you are not aiming that then you can safely let Mr Bayes rest in peace in his tomb and also ignore topics like conditional probability and reverse probability.
:)
TRUE, but we cannot openly suggest the common GMAT aspirants to know or use the terms and names that are not introduced to them until their 10th standard mathematics. We can however make them think (may be) more than the NOT introduced concepts using their common wits that they could have had until their 10th standard mathematics.
Then you should not have posted this question here as even with common wits,it would need only genius few in a class 10th standard to understand the nuances behind the concept of conditional probability.
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by sanju09 » Mon Feb 22, 2010 3:51 am
firdaus117 wrote:
sanju09 wrote:
firdaus117 wrote:
sanju09 wrote: I am checking if the thinkers can deduce something like that theorem on their own, without knowing or naming Mr Bayes, if impossible, kindly ignore the sum for now.
Dude,the questions on probability are generally encountered by the people who score 49-51 in GMAT Maths.If you are not aiming that then you can safely let Mr Bayes rest in peace in his tomb and also ignore topics like conditional probability and reverse probability.
:)
TRUE, but we cannot openly suggest the common GMAT aspirants to know or use the terms and names that are not introduced to them until their 10th standard mathematics. We can however make them think (may be) more than the NOT introduced concepts using their common wits that they could have had until their 10th standard mathematics.
Then you should not have posted this question here as even with common wits,it would need only genius few in a class 10th standard to understand the nuances behind the concept of conditional probability.
...and the GMAT always look for those FEW
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by ajith » Mon Feb 22, 2010 4:37 am
firdaus117 wrote:
ajith wrote:
sanju09 wrote:A lot contains 20 articles. The probability that the lot contains exactly 2 defective articles is 0.4; and the probability that it contains exactly 3 defective articles is 0.6. Articles are drawn from the lot at random, one by one, without replacement until all the defective articles are found. What is the probability that the testing procedure stops at the twelfth testing?

Choices are under construction, tough lot, please bear.
The lot either contains 2 defectives or 3 defectives with probability 0.4 and 0.6 respectively

Say the lot contains 3 defectives, the probability that all of the three would be selected in first 12 = 17C9*3C3/20C12 =0.192

Say the lot contains 2 defectives, the probability that all of the three would be selected in first 12 = 18C10*2C2/20C12 =0.347


probability that the testing procedure stops at the twelfth testing = 0.6*.192 +0.4*.347 = 0.254
Hey Ajith,
I beg to differ from your solution.In your solution,you have calculated probability of getting 2 defective articles in 12 attempts,it may be in any two from 1 to 12th attempt which is not true.But as soon as 2nd defective item is to be found it has to be stopped here which accd to question is in the twelfth attempt.Similarly for 3 defective items......
Rather you should find 1 defective article in 11 attempts and in twelfth the second defective item.That is how I have done it.
Agree

I will try to adapt the solution now..

Say the lot contains 3 defectives, the probability that two would be selected in first 11 and third in 12th = 17C9*2C2/20C11*1/9 =0.0161
Say the lot contains 2 defectives, the probability that one would be selected in first 11 and second in12th= 18C10*2C2/20C11*1/9 =0.0289

Total probability = (0.6 * 0.0161) + (0.4 * 0.0289) = 0.02122
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by firdaus117 » Mon Feb 22, 2010 6:00 am
ajith wrote: Agree

I will try to adapt the solution now..

Say the lot contains 3 defectives, the probability that two would be selected in first 11 and third in 12th = 17C9*2C2/20C11*1/9 =0.0161
Say the lot contains 2 defectives, the probability that one would be selected in first 11 and second in12th= 18C10*2C2/20C11*1/9 =0.0289

Total probability = (0.6 * 0.0161) + (0.4 * 0.0289) = 0.02122
The first one should be 3C2 and second one 2C1.
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by Aman verma » Mon Feb 22, 2010 7:00 am
[quote="firdaus117"][quote="ajith"]
Agree

I will try to adapt the solution now..

Say the lot contains 3 defectives, the probability that two would be selected in first 11 and third in 12th = 17C9*[color=red]2C2[/color]/20C11*1/9 =0.0161
Say the lot contains 2 defectives, the probability that one would be selected in first 11 and second in12th= 18C10*[color=red]2C2[/color]/20C11*1/9 =0.0289

Total probability = (0.6 * 0.0161) + (0.4 * 0.0289) = 0.02122[/quote]
The first one should be 3C2 and second one 2C1.[/quote]

Yes, the first one should be 3C2 and the second one 2C1. Therefore continuing with Ajith's and Firdaus117 solution,
we get : 0.6X(17C9X3C2)/20C11 X1/9 + 0.4X (18C10X2C1)/20C11X1/9. Ajith & Firdaus you both are the real champs.
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