Question

The probability distribution of a discrete r.v. \( X \) is: \[ \begin{array}{c|ccccc} X = x & 0 & 1 & 2 & 3 & 4 \\ P(X = x) & k & 2k & 4k & 2k & k \\ \end{array} \] Then, the value of \( P(X \leq 2) \) is:

• A. \( \frac{1}{10} \)
• B. \( \frac{7}{10} \)
• C. \( \frac{3}{10} \)
• D. \( \frac{9}{10} \)

Hint:

For discrete probability distributions, always ensure that the sum of the probabilities equals 1 and use this to find unknowns like \( k \).

Answer 1

The Correct Option is B

Step 1: Sum of probabilities.
The sum of all probabilities must equal 1: \[ P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1 \] Substitute the probabilities: \[ k + 2k + 4k + 2k + k = 1 \] Simplifying: \[ 10k = 1 \quad \Rightarrow \quad k = \frac{1}{10} \] Step 2: Calculate \( P(X \leq 2) \).
We are asked to find \( P(X \leq 2) \), which is the sum of \( P(X = 0) \), \( P(X = 1) \), and \( P(X = 2) \): \[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = k + 2k + 4k = 7k \] Substitute \( k = \frac{1}{10} \): \[ P(X \leq 2) = 7 \times \frac{1}{10} = \frac{7}{10} \] Step 3: Conclusion.
Thus, the value of \( P(X \leq 2) \) is \( \boxed{\frac{7}{10}} \).