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Two dice are rolled together. Find the probability of getting:
(i) a multiple of $2$ on one and a multiple of $3$ on the other die.
(ii) the product of two numbers on the top of the two dice is a perfect square number.
(i) a multiple of $2$ on one and a multiple of $3$ on the other die.
(ii) the product of two numbers on the top of the two dice is a perfect square number.
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Total outcomes = $36$
(i) $(2, 3), (2, 6), (3, 2), (3, 4), (3, 6), (4, 3), (4, 6), (6, 2), (6, 3), (6, 4), (6, 6)$
Number of outcomes having multiple of $2$ on one die and a multiple of $3$ on other die = $11$
Hence, $P(E) = \frac{11}{36}$
(ii) $(1, 1), (2, 2), (3, 3), (1, 4), (4, 1), (4, 4), (5, 5), (6, 6)$
Number of outcomes having product of two numbers on the top of the dice is a perfect square number = $8$
$P(E) = \frac{8}{36}$ or $\frac{2}{9}$
(i) $(2, 3), (2, 6), (3, 2), (3, 4), (3, 6), (4, 3), (4, 6), (6, 2), (6, 3), (6, 4), (6, 6)$
Number of outcomes having multiple of $2$ on one die and a multiple of $3$ on other die = $11$
Hence, $P(E) = \frac{11}{36}$
(ii) $(1, 1), (2, 2), (3, 3), (1, 4), (4, 1), (4, 4), (5, 5), (6, 6)$
Number of outcomes having product of two numbers on the top of the dice is a perfect square number = $8$
$P(E) = \frac{8}{36}$ or $\frac{2}{9}$