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A six faced die is so biased that it is twice as likely to show an odd number as an even number when rolled. The probability that the sum of the numbers on the upturned faces is even when the die is thrown twice is

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$\frac{4}{9}$

$\frac{5}{9}$

$\frac{1}{2}$

answer is A.

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Detailed Solution

$P (e) = 2 P (0) \Rightarrow P (e) = \frac{2}{3}, P (0) = \frac{1}{3}$

$E =$ Sum is even when two dice are drawn.

$P (E) = \frac{2}{3} \times \frac{2}{3} + \frac{1}{3} \times \frac{1}{3} = \frac{5}{9}$

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