Question

A die is constructed so that when it is thrown, each of the three even numbers 2, 4 and 6 is twice as likely to come up as each of the odd outcomes 1, 3 and 5. What is the probability that 4 comes up when the die is thrown once?

• A. \( \frac{4}{9} \)
• B. \( \frac{2}{3} \)
• C. \( \frac{2}{9} \)
• D. \( \frac{1}{3} \)

Hint:

When probabilities are not equally likely, first determine the probability of each outcome and then calculate the probability of the desired event.

Answer 1

The Correct Option is C

The total number of outcomes on a fair die is 6 (1, 2, 3, 4, 5, 6). The die is constructed such that the even numbers 2, 4, and 6 are twice as likely to come up as the odd numbers 1, 3, and 5. Let the probability of the odd numbers 1, 3, and 5 be \( p \). Then, the probability of the even numbers 2, 4, and 6 is \( 2p \). Thus, the total probability is: \[ 3p + 3(2p) = 1 \] \[ 3p + 6p = 1 \quad \Rightarrow \quad 9p = 1 \quad \Rightarrow \quad p = \frac{1}{9} \] Therefore, the probability of rolling a 4, which is one of the even numbers, is \( 2p \): \[ P(4) = 2 \times \frac{1}{9} = \frac{2}{9} \]