If a dice is thrown twice, it has a total of (6 × 6) 36 possible outcomes.

If S represents the sample space then,

n(S) = 36

Let A represent events the event such that digit greater than 3 comes in the second throw.

∴ A = {(1,4), (2,4), (3,4), (4,4), (5,4), (6,4), (1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (1,6), (2,6), (3,6), (4,6), (5,6), (6,6)}

⇒ P(A) =

Let B represent events the event such digit greater than 3 comes in the first throw.

Similarly 18 outcomes are possible as were present for event A

⇒ P(B) =

Clearly (4,4), (5,4), (6,4), (4,5), (5,5), (6,5) (4,6), (5,6) and (6,6) are common in both events-

∴ P(A ∩ B) =

We need to find the probability of event such that at least one of the 2 throws give digit greater than 33 i.e. P(A or B) = P(A ∪ B)

Note: By definition of P(E or F) under axiomatic approach(also called addition theorem) we know that:

P(E ∪ F) = P(E) + P(F) – P(E ∩ F)

∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ P(A ∪ B) =

Hence,

P(at least one of the two throws shows digit >3) =

∴ Option (b) is correct choice.