A lot contains 20 articles. The probability that the lot

Question

A lot contains 20 articles. The probability that the lot contains exactly 2 defective articles is 0.4 and the probability that the lot contains exactly 3 defective articles is 0.6. Articles are drawn from the lot at random one by one without replacement and are tested till all defective articles are found. What is the probability that the testing procedure ends at the twelfth testing?

A. 0.24

B. 99/190

C. 6/25

D. 8/55

E. 99/1900

Vithal · Answer

it is without replacement:

assume that you have 2 defective -- 0.4

therefore the probability =
18/20*17/19*16/18*...*2/9*0.4

for 3 items it is 17/20*16/19*....3/9*0.6

Add both to get the answer. I am getting it as 12/20*19 = 2/950

but the answer is not one of the options - I guess I am doing something wrong

sparky · Answer

Ok, let's try this one. prob of picking 2, if the lot contains 2, on the 12th testing is 2*11!*18C10/20P12 = 2*11/(19*20) prob of picking 3, if the lot contains 3, on the 12th testing is 3*11!*17C9/20P12 = 3*11/(19*20) * 5/9 so far that's what we've been doing all the time now Sparky needs to open his probability book :help2

P(A) = P(A|B)P(B) + P(A|C)P(C), where P(B or C) = 1 so P(12) = P(12|2)P(2) + P(12|3)P(3) = 2*11/(19*20) * 4/10 + 3*11/(19*20) * 5/9 * 6/10 = 99/1900, or E :rotate

p.s. Vithal, where do you get this kind of problems? I need to go through my all undegrad statisitics course to figure out the solution. oh, man

Vithal · Answer

was sent by one of my friends - I guess this is from another forum!

can you please explain 17C9 and 18C10 ?

I know for sure that this kind of problem is not going to hit me from scoring 51 though :roll:

HIMALAYA · Answer

do not worry about these kinds of problem. :lol:

you find very easy math problems in real test and you can easily hit 51. but you have to be extra careful. 8-)

sparky · Answer

I will explain for 3, same holds for 2

3*11!*17C9/20P12

denominator is clear - drawing 12, order matters

the 12th element needs to be defective and to be the third defective element. the 12th element has three candidates, therefore multiplication by3.

11! should be clear - mixing up the previous 11 elements.

17C9 means all possible combinations of non-defective elements to fill the sequence.

total non defective - 17, need 9 (12-3) non-defective to fill in the sequence with three defective

gmat2me2 · Answer

Can this be taken as 

Taking the 3 defective pieces


20C12*(0.6)^3*(0.4)^9


20 C12 is obvious 

(0.6) is the probabbility of the defective one and 0.4 is the probability of the non-defective one.......

HowManyToGo · Answer

Let, given an item the probability that it is defective be p
then 
p^2 * (1-p)^18 *20c2 = 0.4
p^3 * (1-p)^17 *20c3 = 0.6,

Hence p is 0.2

The probability that the testing would end in the 12 testing is
1 - ( probability that the lot has atleast 13 defects)

= 1 -  

I would proceed if I am no the right track.

HMTG.

Vithal · Answer

OA is E (Sparky is almost always correct on math)

A lot contains 20 articles. The probability that the lot

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