1.

A die is thrown twice. What is the probability that at least one of the two throws comes up with the number 3 ?(a) \(\frac{11}{12}\)(b) \(\frac{11}{36}\)(c) \(\frac{7}{12}\)(d) \(\frac{1}{6}\)

Answer»

(b) \(\frac{11}{36}\)

Let S = total ways in which two dice can be rolled 

⇒ n(S) = 6 × 6 = 36 

Let A : Event of throwing 3 with 1st dice, 

B : Event of throwing 3 with 2nd dice. 

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

⇒ n(A) = 6 

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

A ∩ B = {(3, 3)} ⇒ n(A ∩ B) = 1 

∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{6}{36}\), P(B) = \(\frac{n(B)}{n(S)}\) = \(\frac{6}{36}\), 

P (A ∩ B) = \(\frac{n(A\cap{B})}{n(S)}\) = \(\frac{1}{36}\)

∴ Required probability = P(Throwing a 3 with at least one of the dice)

= P(A) + P(B) – P(A ∩ B)

= \(\frac{6}{36}\) + \(\frac{6}{36}\) - \(\frac{1}{36}\) = \(\frac{11}{36}\).


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