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:

2/3

The correct answer is Option (3) → 2/3

Let probability of odd number = $p$, then probability of even number = $2p$

Total probability for a die: 3 odd + 3 even → $3p + 3*2p = 3p + 6p = 9p = 1 \Rightarrow p = 1/9$

So, $P(\text{odd}) = 1/9$, $P(\text{even}) = 2/9$

Perfect squares on a die: 1,4 → odd:1, even:4

Probability of perfect square in a single toss:

$P(\text{1}) = 1/9$, $P(\text{4}) = 2/9$

$P(\text{perfect square}) = 1/9 + 2/9 = 3/9 = 1/3$

Let $X$ = number of perfect squares in 2 tosses → $X \sim \text{Binomial}(n=2, p=1/3)$

Mean of $X$: $E[X] = n \cdot p = 2 \cdot 1/3 = 2/3$

Mean of number of perfect squares = 2/3