Source: Official Guide for GMAT Review 2016 Problem Solving; #136

3

In a box of 12 pens, a total

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?

1 Explanation

2

Javier Jimenez

When I used the standard probability notation P(a) = (9/12), given that 3 are defective out of 12, so once your first event has occurred, then p(b) = (8/11) from the remaining. So P(a and b) = P (a) x p(b)= (9/12) (8/11) = 6/11.

However, i've seen an alternative approach which I do not comprehend and seek to understand. In one of the forum I see people following this approach: Required probability that none of the chosen is defective = 9C2 / 12C2 = 36 / 66 = 6/11...Can someone explain how 9C2 / 12C12 derives to 36 /66? Is there a lesson in the videos that this approach is explained?

Dec 19, 2016 • Comment

Sam Kinsman, Magoosh Tutor

Hi Javier,

Great question! First, it's worth noting that both approaches are right: the correct answer is 6/11.

The second approach uses a counting technique called combinations. We can use combinations to calculate the probability of something using the following fraction:

(# of outcomes that are considered a success) / (# of possible outcomes)

In this problem, a success is choosing 2 pens that are not defective. So we have:

(# of ways of choosing 2 non-defective pens) / (total # of ways of choosing 2 pens)

The first part (the numerator) can be written as 9C2 ("9 choose 2") because we are choosing 2 pens from 9 non-defective ones. The denominator is 12C2 ("12 choose 2") because we are choosing 2 pens from a total of 9. This is called combinations.

If you are not familiar with combinations, I'd suggest watching this lesson video:

https://gmat.magoosh.com/lessons/341-combinations

Then you can watch the following two videos, which explain how we can use combinations to solve probability problems:

https://gmat.magoosh.com/lessons/1032-using-counting-techniques

https://gmat.magoosh.com/lessons/1033-listing-vs-counting-vs-probability-rules

I hope this helps!

Dec 19, 2016 • Reply

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