Question

If a random variable X has the following probability distribution of X

\[ \begin{array}{|l|c|c|c|c|c|c|c|c|} \hline \mathrm{X}=x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \mathrm{P}(\mathrm{X}=x) & 0 & \mathrm{k} & 2 \mathrm{k} & 2 \mathrm{k} & 3 \mathrm{k} & \mathrm{k}^2 & 2 \mathrm{k}^2 & 7 \mathrm{k}^2+\mathrm{k} \\ \hline \end{array} \]

Then \(P(x \geq 6)=\)

• A. \(\frac{19}{100}\)
• B. \(\frac{81}{100}\)
• C. \(\frac{9}{100}\)
• D. \(\frac{91}{100}\)

Hint:

Use the fact that the probabilities add to 1 to form a quadratic in k, choose the valid root and sum P(6) and P(7).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The sum of all probabilities is \(1\), and each probability must be non-negative.

Step 2: Form the equation
\(0+k+2k+2k+3k+k^2+2k^2+7k^2+k=1\Rightarrow10k^2+9k-1=0\).

Step 3: Solve
\((10k-1)(k+1)=0\Rightarrow k=\frac1{10}\) (the root \(k=-1\) would give negative probabilities).

Step 4: Calculation
\(P(X\ge6)=2k^2+7k^2+k=9k^2+k=\frac9{100}+\frac{10}{100}=\frac{19}{100}\), option (A).

Final Answer:
P(X at least 6) equals \(\frac{19}{100}\). \[ \boxed{\text{(A) }\dfrac{19}{100}} \]