Question

The probability distribution of a random variable \( X \) is given below:
\[ \begin{array}{|c|c|c|c|c|c|} \hline X = x & 10 & 20 & 30 & 40 & 50 \\ \hline P(X = x) & k & 2k & 3k & 4k & 5k \\ \hline \end{array} \] Then, \( P(X = 50) - \dfrac{P(X < 30)}{P(X > 20)} \) = ...........

• A. \(\frac{2}{3}\)
• B. \(\frac{5}{6}\)
• C. \(\frac{1}{12}\)
• D. 0

Hint:

Always ensure the total probability equals 1 to solve for constants in probability distributions.

Answer 1

The Correct Option is C

Step 1: Total probability must be 1
\[ P(X=10) + P(X=20) + P(X=30) + P(X=40) + P(X=50) = 1 \\ k + 2k + 3k + 4k + 5k = 15k = 1 \Rightarrow k = \frac{1}{15} \] Step 2: Evaluate each required probability:
\[ P(X = 50) = 5k = 5 \times \frac{1}{15} = \frac{1}{3} \] \[ P(X < 30) = P(X = 10) + P(X = 20) = k + 2k = 3k = \frac{3}{15} = \frac{1}{5} \] \[ P(X > 20) = P(X = 30) + P(X = 40) + P(X = 50) = 3k + 4k + 5k = 12k = \frac{12}{15} = \frac{4}{5} \] Step 3: Apply the expression:
\[ P(X = 50) - \frac{P(X < 30)}{P(X > 20)} = \frac{1}{3} - \frac{\frac{1}{5}}{\frac{4}{5}} = \frac{1}{3} - \frac{1}{4} = \frac{4 - 3}{12} = \frac{1}{12} \] Final Answer: \( \boxed{\frac{1}{12}} \)