I also feel the same as you guys.

Even though we get stuck, we can still find the right answer.
If 3 disks are to be picked up randomly, the probability would be 3/14.
By adding additional, the probability to chose 3 disks with 8 as a median has to be less than 3/14.
Fortunately, only one answer fulfills that condition.
So, the answer is A.

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) = 42

Now 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) = 364

So, P(median is 8) = 42/364 = 3/26

Answer: A

Cheers,
Brent

Let us assume that we have to pick A, B and C out of these 14 integers. Since, the median has to be 8, B must be 8.

For A we have 7 options (1, 2, 3, 4, 5, 6 and 7)
For C we have 6 options (9, 10, 11, 12, 13 and 14)

Possible ways of picking ABC = 7 x 1 x 6 = 42 (Note: 1 is for B=8, we cannot place any other integer here)

Total number of ways of picking 3 numbers from 14 = 14C3 = 364

Hence, required prob. = 42/364 = 3/26, Option A

here we need one 8, other between 1-7 and other 9-14
small median and large
cases: SML
SLM
MLS
MSL
LSM
LMS
1 case : 7/14*1/13*6/12=1/13*4
2 case will yield same outcome so case 1 and 2 combine 1/26
34 gives 1/26 56 gives 1/26
therefore adding 3/26