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(D) Let $P(O)$ be the probability of getting an odd number and $P(E)$ be the probability of getting an even number.Since the die is biased such that a

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$\frac{5}{9}$

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(D) Let $P(O)$ be the probability of getting an odd number and $P(E)$ be the probability of getting an even number.Since the die is biased such that an even number is twice as likely as an odd number, we have $P(E) = 2P(O)$.Since the sum of probabilities is $1$, $P(E) + P(O) = 1$, which gives $3P(O) = 1$, so $P(O) = \frac{1}{3}$ and $P(E) = \frac{2}{3}$.The sum of two numbers is even if both are even or both are odd.$P(	ext{Sum is even}) = P(E, E) + P(O, O)$.$P(E, E) = P(E) 	imes P(E) = \frac{2}{3} 	imes \frac{2}{3} = \frac{4}{9}$.$P(O, O) = P(O) 	imes P(O) = \frac{1}{3} 	imes \frac{1}{3} = \frac{1}{9}$.Therefore, $P(	ext{Sum is even}) = \frac{4}{9} + \frac{1}{9} = \frac{5}{9}$.

$A$ six-faced die is biased such that it is twice as likely to show an even number as an odd number when thrown. If it is thrown twice, what is the probability that the sum of the two numbers obtained is even?

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