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) = n (A) / n (S)
= 6 /36
= 1/6
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) = n (B) / n (S)
= 6 /36
= 1/6
It is clear that (3,3) is common in both events so,
P (A ∩ B) = n (A ∩ B) / n (S)
= 1/ 36
Now 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)
By using the 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)
So, P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
= 1/6 + 1/6 – 1/36
= 1/3 – 1/36
= 11/36
∴ P (at least one of the two throws comes to be 3) is 11/36
