Question

A random variable \(X\) has the following probability distribution : 

x	0	1	2	3	4
P(x)	k	2k	4k	6k	8k

The value of \(P(1 < X < 4 | x ≤ 2)\) is equal to

• A. \(\frac{4}{7}\)
• B. \(\frac{2}{3}\)
• C. \(\frac{3}{7}\)
• D. \(\frac{4}{5}\)

Answer 1

The Correct Option is A

To solve the given problem, we need to find the value of \( P(1 < X < 4 | x \leq 2) \). This is a conditional probability question. 

The probability distribution of the random variable \( X \) is given by:

x	0	1	2	3	4
P(x)	k	2k	4k	6k	8k

Firstly, since these are probabilities, the sum of all probabilities should be equal to 1:

\[ k + 2k + 4k + 6k + 8k = 1 \] \[ 21k = 1 \] \[ k = \frac{1}{21} \]

Now, we calculate \( P(1 < X < 4) \), which means we need to find the probability of \( X = 2 \) or \( X = 3 \).

\[ P(1 < X < 4) = P(X = 2) + P(X = 3) \] \[ P(1 < X < 4) = 4k + 6k = 10k \] \[ P(1 < X < 4) = 10 \times \frac{1}{21} = \frac{10}{21} \]

Next, we calculate \( P(x \leq 2) \), which means finding the probability of \( X = 0 \), \( X = 1 \), or \( X = 2 \).

\[ P(x \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \] \[ P(x \leq 2) = k + 2k + 4k = 7k \] \[ P(x \leq 2) = 7 \times \frac{1}{21} = \frac{7}{21} = \frac{1}{3} \]

The conditional probability \( P(1 < X < 4 | x \leq 2) \) is calculated as:

\[ P(1 < X < 4 | x \leq 2) = \frac{P(1 < X < 4 \cap x \leq 2)}{P(x \leq 2)} \]

Only \( X = 2 \) falls into both \( 1 < X < 4 \) and \( x \leq 2 \). So:

\[ P(1 < X < 4 \cap x \leq 2) = P(X = 2) = 4k = \frac{4}{21} \]

Substitute this back in to find the conditional probability:

\[ P(1 < X < 4 | x \leq 2) = \frac{\frac{4}{21}}{\frac{1}{3}} = \frac{4}{21} \times \frac{3}{1} = \frac{4}{7} \]

Hence, the value of the conditional probability is \( \frac{4}{7} \).