Question

The probability distribution of a random variable is given below: \[\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline X = x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline P(X = x) & 0 & K & 2K & 2K & 3K & K^2 & 2K^2 & 7K^2 + K \\ \hline \end{array}\]

Find \( P(0<X<5) \).

• A. \( \frac{1}{10} \)
• B. \( \frac{3}{10} \)
• C. \( \frac{8}{10} \)
• D. \( \frac{7}{10} \)

Hint:

To solve probability distribution problems, always ensure the total probability sums to 1, then solve for unknown constants accordingly.

Answer 1

The Correct Option is C

Step 1: Compute Total Probability
The sum of all probabilities must be equal to 1: \[ 0 + K + 2K + 2K + 3K + K^2 + 2K^2 + (7K^2 + K) = 1 \] \[ 9K + 10K^2 = 1 \] \[ 10K^2 + 9K - 1 = 0 \] \[ 10K^2 + 10K - K - 1 = 0 \] \[ 10K(K + 1) - 1(K + 1) = 0 \] \[ (K + 1)(10K - 1) = 0 \]
Step 2: Solve for \( K \) \[ K = -1, \quad K = \frac{1}{10} \] Since probability cannot be negative, we take \( K = \frac{1}{10} \).
Step 3: Compute \( P(0<X<5) \) \[ P(0<X<5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \] \[ = K + 2K + 2K + 3K = 8K \] \[ = 8 \times \frac{1}{10} = \frac{8}{10} \] Final Answer: The correct answer is \(\boxed{(c) \frac{8}{10}}\).