In a box of 12 pens, a total of 3 are defective. If a customer buys 2

My Solution:

Total Pens = 12 Nos

Defective pens = 3

Remaining = 9, Therefore, selecting 2 no defective pens from 9 = 9C2 ways, and selecting 2 pens from 12 = 12C2 ways

Probability of neither pen will be defective = 9C2/12C2 = 36/66 = 6/11 Answer is C

Total number of pens = 12
Defective pens = 3
Non- defective pens= 9

Probablity of selecting 2 non defective pens = 9C2/ 12C2 = 6/11
Answer C

Alternatively , we can use probablity = No of favorable outcomes / No of total outcomes
= 9/12 * 8/11 = 6/11

Bunuel What if Probability (Neither pen will be defective) = 1-Probability (defective both time) = 1- (3/12)*(2/11) = 21/22 ?

I get the different yield.

P(Neither pen will be defective) =

= 1 - (P(both pens are defective) + P(one of the pens is defective)) =

= 1 - (3/12*2/11 + 2*3/12*9/11) =

= 6/11.

Hope it's clear.­

How do you know when to use the (9/12) * (8/11) and when to use (9/12) * (9/12)? It doesn't explicitly say that the customer picks one pen and then another pen afterward. Couldn't the customer have taken both pens at the same time?

1. If the drawing is with replacement it's explicitly mentioned. If it's not mentioned, then it's without replacement.

2. Mathematically the probability of picking two balls simultaneously, or picking them one at a time (without replacement) is the same.

Since there are 3 defective pens from 12, the probability of selecting the first non-defective pen is 9/12 and the probability of selecting the second non-defective pen is 8/11. Thus, the probability that a customer buys 2 non-defective pens is 9/12 x 8/11 = 3/4 x 8/11 = 24/44 = 6/11.

Answer: C

P(neither pen is defective) = P(1st pen selected is NOT defective AND 2nd pen selected is NOT defective)
= P(1st pen selected is NOT defective) x P(2nd pen selected is NOT defective)
= 9/12 x 8/11
= 6/11
= C

ASIDE: How did I get 9/12 and 8/11?
For the first selection, 9 of the 12 pens are good.
For the second selection, we must assume that the first selection resulted in a GOOD pen. This means there are now 11 pens remaining, and 8 of them are GOOD.

hi can some kindly point out my mstake
3c2/12c2+(3c1*9c1/12c2)2

I understand that it should be 3c1/12*9c1/11 but how should i distinguish them. like which one to use where Kindly guide me
Zillion thanks in advance­

Question:
In a box of 12 pens, a total of 3 are defective. If a customer buys 2 pens selected at random from the box, what is the probability that neither pen will be defective?

There are 9 non-defective pens out of 12

If you want to pick simultaneously VS you pick one at a time without replacement, the answers would be SAME

If you want to do this by picking simultaneously, you should interpret the question as:
Favorable: Pick 2 pens that are not defective i.e. pick 2 pens from 9 = 9C2 ways = 9*8/2! = 36 ways
Total: Pick 2 pens from 12 = 12C2 ways = 12*11/2! = 66 ways
=> Probability = 36/66 = 6/11

If you want to do this by picking one at a time, you should interpret the question as:
Pick 1st pen which is not defective AND Pick 2nd pen which is not defective (Note: AND means MULTIPLY)

Probability of picking 1st pen which is not defective = 9/12

Not, since we pick WITHOUT REPLACEMENT, we have 8 non-defective out of 11

Thus, probability of picking 2nd pen which is not defective = 8/11

Thus, required probability = 9/12 * 8/11 = 6/11­

Your reasoning is not correct. Suppose you are choosing 2 pencils out of 9 pencils which are all good; i.e. there are no defective pencils in the box. The two pencils you choose will certainly be both non-defective. In this case, the probability that neither pencil is defective is 1 or equivalently, 100%. According to your reasoning, since there are 9 good pencils and we are choosing 2, the probability should have been 2/9.
Conversely, suppose that there are like a million defective pencils in the box and only 9 good pencils. Again according to your method, since there are 9 good pencils and we are choosing 2, the probability should have been 2/9 but in reality, you can observe that if there were a million defective pencils and only 9 good pencils, the probability of choosing two good pencils is very very low, almost zero.

For the other question you mentioned, if there were 5 red and 4 blue balls in a box and we were choosing 2 balls without replacement; the probability that both are red is (5/9)*(4/8) = 5/18 and the probability that both are blue is (4/9)*(3/8) = 1/6.

Answer: Option C

Step-by-Step Video solution by GMATinsight

In a box of 12 pens, a total of 3 are defective. If a customer buys 2 pens selected at random from the box, what is the probability that neither pen will be defective?

In a box of 12 pens, 9 of them are not defective and 3 are defective. We are selecting 2 pens at random.

Total outcomes here is selecting 2 pens out of 12 = 12C2
Favorable outcomes = Selecting 2 pens out of 9 non defective pens = 9 C2

P(neither pen will be defective) = Fav Outcomes/ Total Outcomes = 9C2/12C2 = 9*8/12*11 = 6/11

Option C is the answer.

Thanks,
Clifin J Francis,
GMAT SME

Hi GMATters,

Here's my video solution to this problem:

­
 Why are we multiplying? The question says "neither" of the pens so its using the "OR" condition and so shouldn't we add the probabilities?
 

­"Neither pen will be defective" means that the first pen is not defective AND the second pen is not defective. Therefore, we multiply the probabilities of each event to find the probability that both conditions are satisfied simultaneously.

­Classic probability: