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

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

X	1	2	3	4	5
P(X)	K	2K	2K	3K	K

The sum of all probabilities must equal 1: 

\(K + 2K + 2K + 3K + K = 9K = 1\)

Solving for \(K\):

\(K = \frac{1}{9}\)

To find \(p = P(1<X<4 | X<3)\), use conditional probability:

\(P(1<X<3) = P(X=2) = 2K = \frac{2}{9}\)

\(P(X<3) = P(X=1) + P(X=2) = K + 2K = 3K = \frac{3}{9}\)

Conditional probability is:

\(p = \frac{P(1<X<3)}{P(X<3)} = \frac{\frac{2}{9}}{\frac{3}{9}} = \frac{2}{3}\)

Given \(5p = \lambda K\), substituting the values:

\(5 \times \frac{2}{3} = \lambda \times \frac{1}{9}\)

\(\frac{10}{3} = \frac{\lambda}{9}\)

\(\lambda = 30\)

Confirming the computed value \(\lambda = 30\) falls within the range (30,30).