Question

When an unfair dice is thrown, the probability of getting a number \( k \) on it is \( P(X = k) = k^2 P \), where \( k = 1, 2, 3, 4, 5, 6 \) and \( X \) is the random variable denoting a number on the dice. Then, the mean of \( X \) is:

• A. \( 25 \)
• B. \( 5 \)
• C. \( \frac{441}{9} \)
• D. \( \frac{441}{91} \)

Hint:

For probability problems involving weighted dice, always normalize the probabilities by summing over all possible values to ensure they sum to 1.

Answer 1

The Correct Option is D

Step 1: Define the Expectation Formula The expectation (mean) of a discrete random variable \( X \) is given by: \[ E(X) = \sum k P(X = k) \]

Step 2: Given Probability Distribution We are given that: \[ P(X = k) = k^2 P \] for \( k = 1, 2, 3, 4, 5, 6 \). Since the total probability must sum to 1: \[ \sum_{k=1}^{6} k^2 P = 1 \] \[ P \sum_{k=1}^{6} k^2 = 1 \] Calculating \( \sum k^2 \): \[ \sum_{k=1}^{6} k^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 = 1 + 4 + 9 + 16 + 25 + 36 = 91 \] So, \[ P(91) = 1 \] \[ P = \frac{1}{91} \]
Step 3: Compute \( E(X) \) \[ E(X) = \sum_{k=1}^{6} k (k^2 P) \] \[ E(X) = P \sum_{k=1}^{6} k^3 \] Using the formula for the sum of cubes: \[ \sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2 \] For \( n = 6 \): \[ \sum_{k=1}^{6} k^3 = \left( \frac{6(7)}{2} \right)^2 = \left( \frac{42}{2} \right)^2 = 21^2 = 441 \] \[ E(X) = P \times 441 = \frac{441}{91} \]
Final Answer: \[ \frac{441}{91} \]