Question

The random variable X has a probability distribution P(X) of the following form, where k is some number.

\[P(X=x) \begin{cases} k & \quad \text{if  } x=0\\   2k  & \quad \text{if }  \text{ x=1}\\  3k & \quad \text{if }  \text{ x=2} \\   0  & \quad \text{otherwise}  \end{cases}\]

Then P(x≤2) is:

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

Answer 1

The Correct Option is B

Given the probability distribution of a random variable \(X\):

\[ P(X=x) = \begin{cases} k & \quad \text{if } x=0 \\ 2k & \quad \text{if } x=1 \\ 3k & \quad \text{if } x=2 \\ 0 & \quad \text{otherwise} \end{cases} \]

To find \(P(x \leq 2)\), we need to determine the probabilities of \(X = 0\), \(X = 1\), and \(X = 2\), and then sum them up.

Since these are the only non-zero probabilities, the sum of probabilities is:

\[ P(X = 0) + P(X = 1) + P(X = 2) = k + 2k + 3k \]

This should equal 1 because the sum of all probabilities for a probability distribution is always 1. Therefore, we have:

\[ k + 2k + 3k = 1 \]

\[ 6k = 1 \]

Solving for \(k\):

\[ k = \frac{1}{6} \]

Substitute back to find each probability:

\[ P(X = 0) = k = \frac{1}{6} \]

\[ P(X = 1) = 2k = \frac{2}{6} = \frac{1}{3} \]

\[ P(X = 2) = 3k = \frac{3}{6} = \frac{1}{2} \]

Now sum these probabilities:

\[ P(x \leq 2) = \frac{1}{6} + \frac{1}{3} + \frac{1}{2} \]

To add these fractions, find a common denominator, which is 6:

\[ \frac{1}{6} + \frac{2}{6} + \frac{3}{6} = \frac{6}{6} \]

\[ P(x \leq 2) = 1 \]

Thus, the probability \(P(x \leq 2)\) is 1.