Question

Given that two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of the numbers on the dice is 4’.

Answer 1

When the numbers appearing on throwing two dice are different.

⇒ The numbers are not the same

Total number of exhaustive cases = 36

Number of cases when doublets do not occur = 36 – 6 = 30

Cases when the sum is 4 are {(1, 3), (2, 2), (3, 1)}

Let A denotes the event when the sum of numbers on two dice is 4 and B is the event when the numbers appearing on the dice are different.

∴ A ∩ B = {(1, 3), (3, 1)}

So, P(A ∩ B) = \(\frac{2}{36}\), P(B) = \(\frac{30}{36}\).

∴ P(A/B) = \(\frac{P(A∩B)}{P(B)}\) = \(\frac{2}{36}\) ÷ \(\frac{3}{36}\) = \(\frac{1}{15}\).