According to question bag contains 5+4+3 = 12 balls

∴ 1 ball can be chosen in ¹²C₁ = 12 ways

Let B denotes the event of drawing a black ball and R denotes event of drawing red ball

∴ P(B) = 5/12 and P(R) = 3/12

We have to find P(B ∩ R)

As, we need to find P(B ∪ R)

∵ Black and red balls can’t be drawn simultaneously.

∴ P(B ∩ R) = 0

∴ P(B ∪ R) = P(B) + P(R) = 5/12 + 3/12 = 8/12 = 2/3

Our answer matches with option (d)

∴ Option (d) is the only correct choice