Pot "A" contains 5 blue and 10 green balls and another pot "B" contains 7 blue and 2 green balls. A biased dice with six sides numbered 1 to 6 is rolled. The probability of eah odd out ome is same and the probability of each even outcome is same. But the probability of an odd outcome is twice than the probability of an even outcome. If the face 1 or 2 or 3 comes up, a ball is taken from the pot "A" else a ball is taken from the pot "B". Find the probability of drawing a green ball.

Let the probability of an even outcome of dice be $$'x'$$.

Then the probability of the odd outcome of the dice is $$'2x'$$.

The sum of all probabilities of all events is 1.

It implies $$x+x+x+2x+2x+2x=1$$

$$x=\frac{1}{9}$$

Probability of drawing a green ball from pot 'A' = Probability of selecting pot 'A' x probability of drawing a green ball.

= $$\left(\frac{2}{9}+\frac{1}{9}+\frac{2}{9}\right)\times\ \frac{10}{15}$$ =$$=\frac{5}{9}\times\ \frac{10}{15}=\frac{10}{27}$$

Similarly, the Probability of drawing a green ball from pot 'B' = $$\left(\frac{1}{9}+\frac{2}{9}+\frac{1}{9}\right)\times\ \frac{2}{9}$$

= $$=\frac{4}{9}\times\ \frac{2}{9}=\frac{8}{81}$$

Probability of drawing a green ball = Probability of drawing a green ball from pot 'A' +Probability of drawing a green ball from pot 'B'.

= $$=\frac{10}{27}+\ \frac{8}{81}=\frac{38}{81}$$

Option (A) is the answer.