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Let a die be loaded in such a way that even faces are twice likely to occur as the odd faces. What is the probability that a prime number will show up when the die is tossed?
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The correct answer is:
$\frac{4}{9}$
Possible prime numbers on the dice are 2,3 and 5 .
Probability of getting prime number $=\frac{2}{3} \times \frac{1}{3}+\frac{1}{3} \times \frac{2}{3}$.
$=\frac{2}{9}+\frac{2}{9}$
$=\frac{4}{9} .$
Probability of getting prime number $=\frac{2}{3} \times \frac{1}{3}+\frac{1}{3} \times \frac{2}{3}$.
$=\frac{2}{9}+\frac{2}{9}$
$=\frac{4}{9} .$
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