Question

A biased die has six faces numbered as \( k = 1, 2, 3, 4, 5, 6 \). On rolling the die, the probability of the number \( k \) appearing is proportional to \( k^2 \). What is the probability that an even number will appear on rolling the die?

• A. \( \frac{35}{91} \)
• B. \( \frac{56}{91} \)
• C. \( \frac{12}{21} \)
• D. \( \frac{9}{21} \)

Hint:

When dealing with probabilities proportional to powers of numbers, first calculate the sum of the powers, and then find the probability for the desired outcomes (e.g., even numbers in this case).

Answer 1

The Correct Option is B

Let the probability of number \( k \) appearing be proportional to \( k^2 \). So, the probability of rolling \( k \) is \( P(k) = \frac{k^2}{\sum_{k=1}^6 k^2} \). First, calculate the sum of the squares of the numbers from 1 to 6: \[ \sum_{k=1}^6 k^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 = 1 + 4 + 9 + 16 + 25 + 36 = 91. \] Thus, the probability of rolling an even number (i.e., \( k = 2, 4, 6 \)) is: \[ P(\text{even}) = P(2) + P(4) + P(6) = \frac{2^2}{91} + \frac{4^2}{91} + \frac{6^2}{91} = \frac{4 + 16 + 36}{91} = \frac{56}{91}. \] Therefore, the probability that an even number will appear is \( \frac{56}{91} \).
Final Answer: \[ \boxed{(B) \, \frac{56}{91}.} \]