Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is:
(i) 0
(ii) 12
(iii) less than 12
(iv) less than or equal to 12

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Solution

Total numbers = 36 i.e.
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
(i) Favourable outcomes=0
P(E)= $\frac{0}{36} = 0$
(ii) Favourbale outcomes= 1 i.e. (6,6)
P(E) = $\frac{1}{36}$
(iii) Favourbale outcomes= 35 [Except (6,6)]
P(E) = $\frac{35}{36}$
(iv) Favourbale outcomes= 36
P(E) = $\frac{36}{36} = 1$

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Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is: (i) 0 (ii) 12 (iii) less than 12 (iv) less than or equal to 12

Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is:
(i) 0
(ii) 12
(iii) less than 12
(iv) less than or equal to 12