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: \[ \frac{441}{91} \]

Answer 2

To determine the mean of \( X \), an unfair dice, we start by understanding that the probability of rolling a number \( k \) is given by \( P(X = k) = k^2 P \), where \( k = 1, 2, 3, 4, 5, 6 \). Given this, the probabilities must sum to 1:
\(\sum_{k=1}^{6} P(X=k)=1\).
This implies: \(P(1^2)+P(2^2)+P(3^2)+P(4^2)+P(5^2)+P(6^2)=1\), or:
\((1^2 \cdot P) + (2^2 \cdot P) + (3^2 \cdot P) + (4^2 \cdot P) + (5^2 \cdot P) + (6^2 \cdot P) = 1\).
Simplifying, we get:
\(1P + 4P + 9P + 16P + 25P + 36P = 1\)
\(91P = 1\)
Solving for \(P\), we have \(P = \frac{1}{91}\).
To find the mean \(E(X)\), use:
\(E(X) = \sum_{k=1}^{6} k \cdot P(X=k)\)
\[ E(X) = 1(1^2P) + 2(2^2P) + 3(3^2P) + 4(4^2P) + 5(5^2P) + 6(6^2P) \]
\[ = (1^3 \cdot P) + (2^3 \cdot P) + (3^3 \cdot P) + (4^3 \cdot P) + (5^3 \cdot P) + (6^3 \cdot P) \]
\[ = (1 + 8 + 27 + 64 + 125 + 216)P \]
\[ = 441P \]
Since \(P = \frac{1}{91}\), then:
\[ E(X) = 441 \times \frac{1}{91}\]
\[ = \frac{441}{91}\]
Thus, the mean of \( X \) is \(\frac{441}{91}\).