Question

A random variable has the following probability distribution

\(X=x_i\)	2	3	4	5
\(P(X=x_i)\)	4k	k	5k	2k

The value of P(X <3) is:

• A. \(\frac{1}{12}\)
• B. \(\frac{1}{3}\)
• C. \(\frac{1}{4}\)
• D. \(\frac{5}{12}\)

Answer 1

The Correct Option is B

To find the probability \(P(X < 3)\), we need the cumulative probability of all outcomes where \(X\) is less than 3. Given the probability distribution:

\(X = x_i\)	2	3	4	5
\(P(X = x_i)\)	4k	k	5k	2k

We calculate \(P(X < 3) = P(X = 2) = 4k\). 

The total probability is the sum of all probabilities:

\[ 4k + k + 5k + 2k = 12k \]

Since the total probability must equal 1, we solve:

\[ 12k = 1 \Rightarrow k = \frac{1}{12} \]

Now substitute the value of \(k\) into \(P(X < 3)\):

\[ P(X < 3) = 4k = 4 \times \frac{1}{12} = \frac{4}{12} = \frac{1}{3} \]

\[ \boxed{P(X < 3) = \frac{1}{3}} \]