Question

Consider the probability distribution

\[ \begin{array}{|l|l|l|l|l|l|} \hline \mathrm{X}=x & 1 & 2 & 3 & 4 & 5 \\ \hline \mathrm{P}(\mathrm{X}=x) & \mathrm{K} & 2 \mathrm{~K} & \mathrm{~K}^2 & 2 \mathrm{~K} & 5 \mathrm{~K}^2 \\ \hline \end{array} \]

Then the value of \(\mathrm{P}(\mathrm{X}>2)\) is

• A. \(\frac{7}{12}\)
• B. \(\frac{1}{36}\)
• C. \(\frac{1}{2}\)
• D. \(\frac{23}{36}\)

Hint:

The probabilities must add to 1, which fixes K; then add the probabilities for X = 3, 4 and 5.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The sum of all probabilities is 1: \(K+2K+K^2+2K+5K^2=1\).

Step 2: Find K
\(6K^2+5K-1=0\Rightarrow(6K-1)(K+1)=0\). Since a probability cannot be negative, \(K=\dfrac16\).

Step 3: Calculation
\(P(X\gt2)=P(3)+P(4)+P(5)=K^2+2K+5K^2=6K^2+2K\)
\(=\dfrac{6}{36}+\dfrac{2}{6}=\dfrac16+\dfrac13=\dfrac12\)

Step 4: Compare the options
This matches option (C).

Final Answer:
P(X > 2) equals 1/2. \[ \boxed{\text{(C) }\dfrac{1}{2}} \]