Question

The probability distribution of a random variable \( X \) is: 
\[\begin{array}{c|c|c|c|c|c} \hline X & 0 & 1 & 2 & 3 & 4 \\ \hline P(X) & 0.1 & k & 2k & k & 0.1 \\ \hline \end{array}\]
 where \( k \) is some unknown constant. The probability that the random variable \( X \) takes the value 2 is: 
 

• A. \( \frac{1}{5} \)
• B. \( \frac{2}{5} \)
• C. \( \frac{4}{5} \)
• D. \( 1 \)

Hint:

To solve probability equations, substitute the known values and simplify step by step. Ensure that the sum of all probabilities equals \( 1 \), as this is the fundamental property of probability distributions.

Answer 1

The Correct Option is B

Step 1: Equation Setup The sum of probabilities is 1: \[ P(0) + P(1) + P(2) + P(3) + P(4) = 1 \] With the given values: \[ 0.1 + k + 2k + k + 0.1 = 1 \] where \( P(1) = k \), \( P(2) = 2k \), and \( P(3) = k \).

Step 2: Equation Simplification Combine like terms: \[ 0.2 + 4k = 1 \] Isolate the term with \( k \): \[ 4k = 0.8 \] Solve for \( k \): \[ k = 0.2 = \frac{1}{5} \]

Step 3: Calculate \( P(2) \) Using the relation \( P(2) = 2k \), substitute the value of \( k \): \[ P(2) = 2 \times \frac{1}{5} = \frac{2}{5} \]

Conclusion: \( P(2) \) equals \( \mathbf{\frac{2}{5}} \).