The probability distribution of a random variable is given below:
`{:(X=x,0,1,2,3,4,5,6,7),(P(X=x),0,k,2k,2k,3k,k^(2),2k^(2),7k^(2)+k):}`
Then `P(0 lt X lt 5)=`

Options

A. `(1)/(10)`
B. `(3)/(10)`
C. `(8)/(10)`
D. `(7)/(10)`

Answer

To solve the problem, we need to find the probability \( P(0 < X < 5) \) given the probability distribution of the random variable \( X \).

### Step 1: Write down the probability distribution
The probability distribution is given as:
- \( P(X=0) = 0 \)
- \( P(X=1) = k \)
- \( P(X=2) = 2k \)
- \( P(X=3) = 2k \)
- \( P(X=4) = 3k \)
- \( P(X=5) = k^2 \)
- \( P(X=6) = 2k^2 \)
- \( P(X=7) = 7k^2 + k \)

### Step 2: Set up the equation for total probability
Since the sum of all probabilities must equal 1, we can write:
\[ 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1 \]
This simplifies to:
\[ (0 + k + 2k + 2k + 3k + k + 2k^2 + 7k^2) = 1 \]
Combining like terms:
\[ 9k + 10k^2 = 1 \]

### Step 3: Rearrange the equation
Rearranging gives us:
\[ 10k^2 + 9k - 1 = 0 \]

### Step 4: Solve the quadratic equation
We can use the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 10, b = 9, c = -1 \):
\[ k = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 10 \cdot (-1)}}{2 \cdot 10} \]
Calculating the discriminant:
\[ 9^2 + 40 = 81 + 40 = 121 \]
So,
\[ k = \frac{-9 \pm 11}{20} \]
Calculating the two possible values:
1. \( k = \frac{2}{20} = \frac{1}{10} \)
2. \( k = \frac{-20}{20} = -1 \) (not valid since probability cannot be negative)

Thus, we have:
\[ k = \frac{1}{10} \]

### Step 5: Calculate \( P(0 < X < 5) \)
Now, we need to calculate \( P(0 < X < 5) \), which is:
\[ P(X=1) + P(X=2) + P(X=3) + P(X=4) \]
Substituting the values:
\[ P(X=1) = k = \frac{1}{10} \]
\[ P(X=2) = 2k = \frac{2}{10} \]
\[ P(X=3) = 2k = \frac{2}{10} \]
\[ P(X=4) = 3k = \frac{3}{10} \]
Adding these probabilities:
\[ P(0 < X < 5) = \frac{1}{10} + \frac{2}{10} + \frac{2}{10} + \frac{3}{10} = \frac{8}{10} = \frac{4}{5} \]

### Final Answer
Thus, the probability \( P(0 < X < 5) = \frac{4}{5} \).

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