Question
Two dice are rolled once. Find the probability of getting such numbers on two dice, whose product is a perfect square.
✓
Answer
When two dice are rolled, then the possible outcomes are:
$
\left(\begin{array}{llllll}
(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)
\end{array}\right)=36 \text { outcomes }
$
Favorable outcomes for the numbers on two dice, whose product is a perfect square are $\{(1,1)(1,4)(2,2)(3,3)(4,1)(4,4)(5,5)(6,6)\}$, i.e. 8 outcomes.
Therefore, probability of getting such numbers on two dice, whose product is a perfect square
$
=\frac{\text { Favourable outcome }}{\text { Total outcome }}=\frac{8}{36}=\frac{2}{9}
$
$
\left(\begin{array}{llllll}
(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)
\end{array}\right)=36 \text { outcomes }
$
Favorable outcomes for the numbers on two dice, whose product is a perfect square are $\{(1,1)(1,4)(2,2)(3,3)(4,1)(4,4)(5,5)(6,6)\}$, i.e. 8 outcomes.
Therefore, probability of getting such numbers on two dice, whose product is a perfect square
$
=\frac{\text { Favourable outcome }}{\text { Total outcome }}=\frac{8}{36}=\frac{2}{9}
$
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