A random variable X has the following distribution
`{:(X=x_(1):,1,2,3,4),(P(X=x_(1)):,k,2k,3k,4k):}`
The value of k and `P(X lt 3)` are equal to

Options

A. `k=(1)/(10), P(x lt 3)=(3)/(5)`
B. `k=(1)/(10), P(x lt 3)=(3)/(10)`
C. `k=(3)/(10), P(x lt 3)=(1)/(10)`
D. `k=(1)/(10), P(x lt 3)=(5)/(12)`

Answer

To solve the problem step by step, we will first determine the value of \( k \) and then calculate \( P(X < 3) \).

### Step 1: Set up the equation for the probabilities
The random variable \( X \) has the following distribution:
- \( P(X = 1) = k \)
- \( P(X = 2) = 2k \)
- \( P(X = 3) = 3k \)
- \( P(X = 4) = 4k \)

Since the sum of all probabilities must equal 1, we can write the equation:
\[ P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1 \]
This translates to:
\[ k + 2k + 3k + 4k = 1 \]

### Step 2: Combine like terms
Combining the terms on the left side gives:
\[ (1 + 2 + 3 + 4)k = 1 \]
\[ 10k = 1 \]

### Step 3: Solve for \( k \)
To find \( k \), we divide both sides by 10:
\[ k = \frac{1}{10} \]

### Step 4: Calculate \( P(X < 3) \)
Now we need to find \( P(X < 3) \), which is the sum of the probabilities for \( X = 1 \) and \( X = 2 \):
\[ P(X < 3) = P(X = 1) + P(X = 2) = k + 2k \]
Substituting the value of \( k \):
\[ P(X < 3) = k + 2k = 3k \]
Now substitute \( k = \frac{1}{10} \):
\[ P(X < 3) = 3 \times \frac{1}{10} = \frac{3}{10} \]

### Final Answers
Thus, the value of \( k \) is \( \frac{1}{10} \) and \( P(X < 3) \) is \( \frac{3}{10} \).