<p>A dice is loaded so that the probability of a face \(i\) is proportional to \(i\), \(i = 1, 2, \ldots, 6\). Then find the probability of an even number occurring when the dice is rolled.</p>
Step-by-Step Solution
Key Concept: Since probability of face i is proportional to i, we have P(i) = ki where k is a constant. Use the fact that all probabilities sum to 1 to find k, then sum probabilities of even faces.
<p><strong>Step 1:</strong> Let P(face i) = ki for i = 1, 2, 3, 4, 5, 6, where k is the proportionality constant.</p><p><strong>Step 2:</strong> Since all probabilities must sum to 1:
$$k(1 + 2 + 3 + 4 + 5 + 6) = 1$$
$$k(21) = 1$$
$$k = \dfrac{1}{21}$$</p><p><strong>Step 3:</strong> The even faces are 2, 4, and 6. Therefore:
$$P(\text{even}) = P(2) + P(4) + P(6)$$
$$= \dfrac{2}{21} + \dfrac{4}{21} + \dfrac{6}{21}$$
$$= \dfrac{12}{21}$$
$$= \dfrac{4}{7}$$</p><p>∴ Answer: $\dfrac{4}{7}$</p>
Correct Answer: \(\dfrac{4}{7}\)