A random variable X has following probability distribution
`{:(X=x,1,2,3,4,5,6),(P(X=x),K,3K,5K,7K,8K,K):}`
Then `P(2leX lt 5)`= . . . .

Options

A. `(3)/(5)`
B. `(7)/(25)`
C. `(23)/(25)`
D. `(24)/(25)`

Explanation

To solve the problem, we need to find the probability \( P(2 \leq X < 5) \) for the given probability distribution of the random variable \( X \).

### Step-by-step Solution:

1. **Identify the Probability Distribution**:
The random variable \( X \) has the following probability distribution:
\[ \begin{align*} X & : 1, 2, 3, 4, 5, 6 \\ P(X=x) & : K, 3K, 5K, 7K, 8K, K \end{align*} \]

2. **Set Up the Equation for Total Probability**:
The sum of all probabilities must equal 1:
\[ K + 3K + 5K + 7K + 8K + K = 1 \]
Simplifying this gives:
\[ 25K = 1 \]

3. **Solve for \( K \)**:
To find \( K \), divide both sides by 25:
\[ K = \frac{1}{25} \]

4. **Calculate Individual Probabilities**:
Now we can calculate the probabilities for each value of \( X \):
\[ \begin{align*} P(X=1) & = K = \frac{1}{25} \\ P(X=2) & = 3K = \frac{3}{25} \\ P(X=3) & = 5K = \frac{5}{25} = \frac{1}{5} \\ P(X=4) & = 7K = \frac{7}{25} \\ P(X=5) & = 8K = \frac{8}{25} \\ P(X=6) & = K = \frac{1}{25} \end{align*} \]

5. **Determine the Required Probability**:
We need to find \( P(2 \leq X < 5) \), which includes \( X = 2, 3, 4 \):
\[ P(2 \leq X < 5) = P(X=2) + P(X=3) + P(X=4) \]
Substituting the values:
\[ P(2 \leq X < 5) = \frac{3}{25} + \frac{5}{25} + \frac{7}{25} \]

6. **Combine the Probabilities**:
Adding these fractions:
\[ P(2 \leq X < 5) = \frac{3 + 5 + 7}{25} = \frac{15}{25} = \frac{3}{5} \]

### Final Answer:
Thus, the probability \( P(2 \leq X < 5) \) is:
\[ \boxed{\frac{3}{5}} \]