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 3 comes in the first throw.

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

⇒ P(A) =

Let B represent events the event such that 3 comes in the second throw.

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

⇒ P(B) =

Clearly (3,3) is 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 3 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 comes to be 3) =