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Total Probability Theorem

A die is load...

Question

A die is loaded so that the probability of a face $i$ is proportional to $i, i = 1, 2, . . .6$ . The probability of an even number occurring when the die is rolled is $\frac{b}{7},$ the value of $b$ is

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Solution

Since the probability of the faces are proportional to the numbers on them, we can take the probabilities of faces $1, 2, . . ., 6$ as $k, 2 k, . . . ., 6 k$ , respectively.
Since one of the faces must occur, we have
$k + 2 k + 3 k + 4 k + 5 k + 6 k = 1 \Rightarrow k = \frac{1}{21}$
Therefore, the probability of occurence of an even number is $2 k + 4 k + 6 k = 12 k = \frac{12}{21} = \frac{4}{7}$

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