The random variable 'X' has a probability distribution P (X) of the following form, where k is some number :
`P(X)={{:(k",","if",X=0),(2k",","if",X=1),(3k",","if",X=2),(0",","if","otherwise"):}`
(a) Determine the value of k.
(b) Find `P(X lt 2),P(X le2),P(X ge2)`

Answer

To solve the given problem, we need to determine the value of \( k \) in the probability distribution and then calculate the required probabilities. Let's break it down step by step.

### Step 1: Determine the value of \( k \)

The probability distribution \( P(X) \) is defined as follows:

- \( P(X = 0) = k \)
- \( P(X = 1) = 2k \)
- \( P(X = 2) = 3k \)
- \( P(X \text{ otherwise}) = 0 \)

Since the sum of all probabilities must equal 1, we can write the equation:

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

Substituting the values we have:

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

Combining like terms gives:

\[ 6k = 1 \]

Now, solving for \( k \):

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

### Step 2: Find the probabilities

Now that we have \( k \), we can substitute it back into the probability distribution:

- \( P(X = 0) = k = \frac{1}{6} \)
- \( P(X = 1) = 2k = \frac{2}{6} = \frac{1}{3} \)
- \( P(X = 2) = 3k = \frac{3}{6} = \frac{1}{2} \)

### Step 3: Calculate \( P(X < 2) \)

To find \( P(X < 2) \), we need to consider the values of \( X \) that are less than 2, which are \( X = 0 \) and \( X = 1 \):

\[ P(X < 2) = P(X = 0) + P(X = 1) = \frac{1}{6} + \frac{1}{3} \]

To add these fractions, we need a common denominator. The least common multiple of 6 and 3 is 6:

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

### Step 4: Calculate \( P(X \leq 2) \)

To find \( P(X \leq 2) \), we include \( X = 0, 1, \) and \( 2 \):

\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = \frac{1}{6} + \frac{1}{3} + \frac{1}{2} \]

Again, we need a common denominator, which is 6:

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

### Step 5: Calculate \( P(X \geq 2) \)

To find \( P(X \geq 2) \), we consider \( X = 2 \) and values greater than 2 (which have a probability of 0):

\[ P(X \geq 2) = P(X = 2) + P(X > 2) = P(X = 2) + 0 = \frac{1}{2} \]

### Summary of Results

(a) The value of \( k \) is \( \frac{1}{6} \).

(b) The probabilities are:
- \( P(X < 2) = \frac{1}{2} \)
- \( P(X \leq 2) = 1 \)
- \( P(X \geq 2) = \frac{1}{2} \)