Question

The probability distribution of random variable X is given by : 

Let \(p = P(1<X<4 | X<3)\). If \(5p = \lambda K\), then \(\lambda\) is equal to _________.
 

Hint:

For conditional probability P(A|B), first clearly identify the outcomes that constitute event A, event B, and their intersection (A \(\cap\) B). Then calculate the probabilities of the intersection and the condition (B), and take their ratio.

Answer 1

Correct Answer: 30

Step 1: Find the value of K.
The sum of all probabilities in a probability distribution must be equal to 1. \[ \sum P(X=x_i) = K + 2K + 2K + 3K + K = 1 \] \[ 9K = 1 \implies K = \frac{1}{9} \] Step 2: Calculate the conditional probability p.
We need to find \(p = P(1<X<4 | X<3)\).
Using the formula for conditional probability, \(P(A|B) = \frac{P(A \cap B)}{P(B)}\).
- Let A be the event \(1<X<4\), which means \(X \in \{2, 3\}\).
- Let B be the event \(X<3\), which means \(X \in \{1, 2\}\).
- The intersection of A and B, \(A \cap B\), is the event where both conditions are true, which is \(X=2\).
Now, we find the probabilities of these events:
- \(P(A \cap B) = P(X=2) = 2K\) - \(P(B) = P(X<3) = P(X=1) + P(X=2) = K + 2K = 3K\) Now calculate p:
\[ p = \frac{P(A \cap B)}{P(B)} = \frac{2K}{3K} = \frac{2}{3} \] Step 3: Use the given relation to find \(\lambda\).
We are given the relation \(5p = \lambda K\).
Substitute the values of p and K that we found:
\[ 5 \left(\frac{2}{3}\right) = \lambda \left(\frac{1}{9}\right) \] \[ \frac{10}{3} = \frac{\lambda}{9} \] Solve for \(\lambda\):
\[ \lambda = \frac{10}{3} \times 9 = 10 \times 3 = 30 \]