A die is rolled in such a way that an even number is twice likely to occur as an odd number. If the die is rolled twice, then the mean of the number of perfect squares in two tosses is:

The die has even numbers that are twice as likely to occur as odd numbers.

Odd numbers on a die: 1, 3, 5

Even numbers on a die: 2, 4, 6

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Let the probability of getting any one odd number = $p$

Then the probability of getting any one even number = $2p$

All probabilities must sum to 1:

$P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1$

$p + 2p + p + 2p + p + 2p = 1$

$9p = 1$

$p = \dfrac{1}{9}$

Therefore:

Probability of each odd number (1, 3, 5) = $\dfrac{1}{9}$

Probability of each even number (2, 4, 6) = $\dfrac{2}{9}$

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Perfect squares between 1 and 6 are: 1 and 4

Probability of getting a perfect square in one toss:

$P(\text{perfect square}) = P(1) + P(4)$

$P(\text{perfect square}) = \dfrac{1}{9} + \dfrac{2}{9}$

$P(\text{perfect square}) = \dfrac{3}{9}$

$P(\text{perfect square}) = \dfrac{1}{3}$

$P(\text{not a perfect square}) = 1 - \dfrac{1}{3} = \dfrac{2}{3}$

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Let $X$ = number of perfect squares obtained in 2 tosses

$X$ can be 0, 1, or 2

$P(X = 0) = \dfrac{2}{3} \times \dfrac{2}{3} = \dfrac{4}{9}$

$P(X = 1) = 2 \times \dfrac{1}{3} \times \dfrac{2}{3} = \dfrac{4}{9}$

$P(X = 2) = \dfrac{1}{3} \times \dfrac{1}{3} = \dfrac{1}{9}$

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Mean = $E[X] = \sum (\text{value} \times \text{probability})$

Mean = $0 \times P(X=0) + 1 \times P(X=1) + 2 \times P(X=2)$

Mean = $0 \times \dfrac{4}{9} + 1 \times \dfrac{4}{9} + 2 \times \dfrac{1}{9}$

Mean = $0 + \dfrac{4}{9} + \dfrac{2}{9}$

Mean = $\dfrac{6}{9}$

Mean = $\dfrac{2}{3}$

Therefore, the mean of the number of perfect squares in two tosses is $\dfrac{2}{3}$.