Asmakan wrote:

The only contents of a container are 14 disks that are each numbered with a different integer that is one of the integers 1 through 14 inclusive. If 3 disks are randomly removed from the container, one after the other, and without replacement, what is the probability that the median of the numbers on the 3 selected disks will be 8?

A) \(\frac{3}{26}\)

B) \(\frac{3}{14}\)

C) \(\frac{2}{7}\)

D) \(\frac{1}{3}\)

E) \(\frac{165}{364}\)

Let's solve this question using counting methods.

In order for the median (of the three disks) to be 8, one of the disks must be 8, one must be less than 8, and the other must be greater than 8.

I how many ways can we have such a configuration?

We can select the 8 in

1 way

We can select a number

less than 8 in

7 ways (1,2,3,4,5,6, or 7)

We can select a number

greater than 8 in

6 ways (9,10,11,12,13 or 14)

By the Fundamental Counting Principle (FCP), the total number of ways to get a median of 8 = (

1)(

7)(

6) =

42Now let's determine the TOTAL number of possible outcomes when we choose 3 discs.

Since the order in which we select the three discs does not matter, we can use COMBINATIONS

We can select 3 of the 14 discs in 14C3 ways

14C3 = (14)(13)(12)/(3)(2)(1) =

364So, P(median is 8) =

42/

364 = 3/26

Answer: A

Cheers,

Brent

_________________

Brent Hanneson – Creator of greenlighttestprep.com

Sign up for my free GRE Question of the Day emails