Question

An assignment of probabilities for outcomes of the sample space \( S = \{1, 2, 3, 4, 5, 6\} \) is as follows:

\[ \begin{array}{c|c c c c c c} 1 & 2 & 3 & 4 & 5 & 6 \\ \hline k & 3k & 5k & 7k & 9k & 11k \end{array} \]

If this assignment is valid, then the value of \( k \) is:

• A. \( \frac{1}{34} \)
• B. \( \frac{1}{35} \)
• C. \( \frac{1}{38} \)
• D. \( \frac{1}{37} \)
• E. \( \frac{1}{36} \)

Hint:

When dealing with probabilities, always remember that the sum of the probabilities for all possible outcomes in a sample space must be equal to 1.

Answer 1

Answer: (E)

For the assignment to be valid, the sum of all probabilities must be equal to 1. 
Therefore, we can write: \[ k + 3k + 5k + 7k + 9k + 11k = 1 \] Simplifying: \[ k(1 + 3 + 5 + 7 + 9 + 11) = 1 \] \[ k \times 36 = 1 \] \[ k = \frac{1}{36} \] Thus, the value of \( k \) is \( \frac{1}{36} \).
Thus, the correct answer is option (E), \( \frac{1}{36} \).