Question

Find the probability that the sum is as stated when a pair of dice is rolled odd OR less than 5

Answer 1

Let's consider two events: the first one - sum of the dices is odd, the second one - sum is less than 5.

In order to get odd sum, the possible combinations can be {1,3,5} (1st die) + {2,4,6} (2nd die) or {2,4,6} (1st die) + {1,3,5} (2nd die). The total number of combinations with odd sum is $3 \times 3 + 3 \times 3 = 18$. There are $6 \times 6$ possible outcomes of throwing 2 dices. So,

$\displaystyle P(A) = \frac{18}{36} = \frac{1}{2}$

To get sum less than 5, we can have pairs (1, 1), (1,2),(2,1), (3,1), (1,3), (2,2). Only 6 combinations satisfy given condition. So,

$\displaystyle P(B) = \frac{6}{36} = \frac{1}{6}$

And finally,

$P(A\, or\, B) = P(A) + P(B) - P(A \, and \, B)$

The last term here is for all events where the sum is odd and less than five (there are only 2 of them - (1,2) and (2,1)). We need to subtract it because otherwise we double-count these events.

$\displaystyle P(A \, or B) = \frac{1}{2} + \frac{1}{6} - \frac{2}{36}=\frac{1}{2} + \frac{1}{6} - \frac{1}{18} = \frac{9+3-1}{18}=\frac{11}{18}$

Answer: $\displaystyle \frac{11}{18}$