Two dice are thrown. Probability that the product is (i) 6 (ii) 12 (iii) 7?
(i) \(\dfrac{1}{9}\), (ii) \(\dfrac{1}{9}\), (iii) \(0\)
Step 1: Total possible outcomes when 2 dice are thrown = \(6 \times 6 = 36\).
Step 2: Case (i) Product = 6.
We need pairs \((a,b)\) such that \(a \times b = 6\).
Possible pairs: (1,6), (6,1), (2,3), (3,2).
So favourable outcomes = 4.
Probability = \(\dfrac{4}{36} = \dfrac{1}{9}\).
Step 3: Case (ii) Product = 12.
We need pairs \((a,b)\) such that \(a \times b = 12\).
Possible pairs: (2,6), (6,2), (3,4), (4,3).
So favourable outcomes = 4.
Probability = \(\dfrac{4}{36} = \dfrac{1}{9}\).
Step 4: Case (iii) Product = 7.
We check all factors of 7. But 7 is a prime number, so only possible product pairs are (1,7) or (7,1).
But dice numbers are only from 1 to 6, so 7 is not possible.
Hence favourable outcomes = 0.
Probability = \(\dfrac{0}{36} = 0\).