The probability that a relation R from {x, y} to {x, y} is both symmetric and transitive, is equal to :

Solution

Maths sir · 5 steps

List the symmetric relations first, there are only 8, then test each one for transitivity.

1. Count all relations. A × A has 4 ordered pairs: (x,x), (y,y), (x,y), (y,x). Each pair is either in R or not, so there are 2⁴ = 16 relations in total.
2. Symmetric ones. (x,x) and (y,y) can be chosen freely. (x,y) and (y,x) must come together or not at all. That gives 2 × 2 × 2 = 8 symmetric relations.
3. Without the cross pair. If R has only diagonal pairs, no chain (a,b), (b,c) can break transitivity. So all 4 of these work: φ, {(x,x)}, {(y,y)}, {(x,x),(y,y)}.
4. With the cross pair. From (x,y) and (y,x), transitivity forces (x,x). From (y,x) and (x,y), it forces (y,y). So only the full relation A × A works here, which gives 1 more.
5. So. There are 4 + 1 = 5 favourable relations out of 16, so the probability is 5/16, option (a).

The idea to keep. On a small set, list the relations with the more restrictive property first, then test the other property. A symmetric pair (a,b), (b,a) together with transitivity forces both (a,a) and (b,b).

Answer from the NTA final answer key. Solution by PrepWiser.