Question

Two dice having six sides with numbers $1,2,3,4,5$ and $6$ written on them or rolled. Answer the questions that follow.
What is the probability that both numbers are odd ?

Answer 1

Let’s figure out the number of possible outcomes. Each has $6$ possible outcomes $(1,2,3,4,5,6)$, so the combination of these two becomes $6\times 6=36$.

Below the first number in the parenthesis is the outcome of the first die and the second number is the outcome of the second die.

Now, of these $36$ possible combinations, pair in which both no. are odd highlighted below ..

$$\begin{gathered} \mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}(1,2)\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{3}\mathbf{)}(1,4)\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{5}\mathbf{)}(1,6) \\ (2,1)(2,2)(2,3)(2,4)(2,5)(2,6) \\ \mathbf{(}\mathbf{3}\mathbf{,}\mathbf{1}\mathbf{)}(3,2)\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{3}\mathbf{)}(3,4)\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{5}\mathbf{)}(3,6) \\ (4,1)(4,2)(4,3)(4,4)(4,5)(4,6) \\ \mathbf{(}\mathbf{5}\mathbf{,}\mathbf{1}\mathbf{)}(5,2)\mathbf{}\mathbf{(}\mathbf{5}\mathbf{,}\mathbf{3}\mathbf{)}(5,4)\mathbf{}\mathbf{(}\mathbf{5}\mathbf{,}\mathbf{5}\mathbf{)}(5,6) \\ (6,1)(6,2)(6,3)(6,4)(6,5)(6,6) \end{gathered}$$

$Probability=\frac{desiredoutcomes}{totaloutcomes}$

Probability = $\frac{9}{36}=\frac{1}{4}$

So, the probability of the given question is $\frac{1}{4}$.