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) = \(\dfrac{ n (A) }{ n (S)}\)

=\( \dfrac{6}{36}\)

= \( \dfrac{1}{6}\)

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) = \( \dfrac{ n (B) }{ n (S)}\)

= \( \dfrac{6}{36}\)

= \( \dfrac{1}{6}\)

It is clear that (3,3) is common in both events so,

P (A ∩ B) = \(\dfrac{ n (A ∩ B) }{ n (S)}\)

= \( \dfrac{1}{36}\)

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)

= \( \dfrac{1}{6} + \dfrac{1}{6} – \dfrac{1}{36}\)

= \( \dfrac{1}{3}- \dfrac{1}{36}\)

= \( \dfrac{11}{36}\)

P (at least one of the two throws comes to be 3) is \( \dfrac{11}{36}\)

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