No, there is not.

Regards

I have two methods for this.

the longer way is:
total number of combinations is 15c2
the total number of ways to select 5 plates is 5c2
the total number of ways to select 3 different sets is 3c2

then
(5c2*3c2)/15c2 = 10*3/105= 30/105= 2/7

the shorter way is..

the probability of selecting any plate is 1
the probability of selecting a matching plate out of the 14 other plates is 4/14
1*4/14=2/7
the answer is A

Another way..
Any of the 15 can be picked up in the first go. so P = \(\frac{15}{15}=1\)
Next can be any of the remaining 10 to ensure they are from different sets so P = \(\frac{10}{14}\)
Over all prob = \(1*\frac{10}{14}=\frac{5}{7}\)

Thus required probability = 1- \frac{5}{7}=\frac{2}{7}[/m]

Solution:

Selecting 2 plates from all the plates ( 3 different sets and each having 5 plates = 3x5 = 15) => 15c2 = (15x14)/2 = 15 x 7

Selecting 2 plates from same set = Selecting 1 set from 3 sets (3c1) and Selected 2 plates from the set selected (5c2).

i.e; 3c1 x 5c2 = 30

Probability = (3c1 x 5c2) / 15c2
= 30/ (15 x 7)
= 2/7

Method 1) Let assume 3 dinner set as AAAAA BBBBB CCCCC

now probability of selecting randomly selecting 2 plates from same set is complementary of Probability of not choosing any plate from same set

P(not choosing any plate from same set) = 5C1 * 5C1/15C2 * 3 (why 3 as first we did this for set A and B , therefore Set B and C , then for Set A and C
or you can simply make combination for 3 dinner set as 3C2)

P(selecting randomly selecting 2 plates from same set) = 1 - 5C1 * 5C1/15C2 * 3
= 1 - 5/7 = 2/7

OR

Method 2) Let assume 3 dinner set as AAAAA BBBBB CCCCC

P(selecting randomly selecting 2 plates from same set) = 5C2/15C2 *3 (why 3 as same probability for all 3 sets)
= 2/7