Two different dice are thrown together. Find the probability that the numbers obtained
(i) have a sum less than 7
(ii) have a product less than 16
(iii) is a doublet of odd numbers.
Total number of possible outcomes is 36.
(i) The favorable outcomes for sum of numbers on dices less than 7 are (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (4,1), (4,2), (5,1).
$P($ the sum of numbers appeared is less than 7$)=\frac{15}{36}=\frac{5}{12}$.
(ii) The favorable outcomes for product of numbers on dices less than 16 are (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), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (6,1), (6,2).
$P($ the product of numbers appeared is less than 16$)=\frac{25}{36}$.
(iii) The favorable outcomes are (1,1), (3,3), (5,5).
$P$ (the doublet of odd numbers) $=\frac{3}{36}=\frac{1}{12}$.
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