Two dice are thrown independently. Let A be the event that the number appeared on the 1ˢᵗ die is less than the number appeared on the 2ⁿᵈ die, B be the event that the number appeared on the 1ˢᵗ die is even and that on the second die is odd, and C be the event that the number appeared on the 1ˢᵗ die is odd and that on the 2ⁿᵈ is even. Then :

Solution

Maths sir · 4 steps

B and C can never happen together, so (A ∪ B) ∩ C shrinks to just A ∩ C. Count that.

1. Count A, B, C. Two dice give 36 outcomes. Of these, 6 are ties, and the other 30 split equally between first smaller and first larger, so n(A) = 15. For B the first die has 3 even choices and the second has 3 odd choices, so n(B) = 3 × 3 = 9. In the same way, n(C) = 9.
2. Simplify (d). Distribute: (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C). The first die cannot be both even and odd, so B ∩ C = ∅. That leaves only A ∩ C: first die odd, second die even, and first smaller than second.
3. Count A and C. First die 1: second can be 2, 4 or 6, which is 3 cases. First die 3: second can be 4 or 6, which is 2 cases. First die 5: second must be 6, which is 1 case. Total 3 + 2 + 1 = 6.

   Option	Check	Verdict
   (a)	(2,3) is in both A and B, so A ∩ B ≠ ∅	false
   (b)	n(B) = n(C) = 9, not 6	false
   (c)	P(B ∩ C) = 0 but P(B) P(C) = 9/36 × 9/36 = 1/16	false
   (d)	n((A ∪ B) ∩ C) = n(A ∩ C) = 6	true

4. So. Only (d) holds: the event (A ∪ B) ∩ C has exactly 6 favourable cases.

The idea to keep. Mutually exclusive and independent are different ideas. Two events that cannot happen together, each with nonzero probability, are always dependent. Use the distributive law to drop empty intersections before counting.

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