To solve the problem, we need to find the probability that the random variable \( X \) takes a prime value, given the probability function \( P(X=k) = a \frac{(k+1)}{2^k} \) for \( k = 0, 1, 2, 3, 4, 5 \).
### Step-by-Step Solution:
1. **Identify Prime Values**:
The prime numbers in the range of \( k = 0, 1, 2, 3, 4, 5 \) are \( 2, 3, \) and \( 5 \).
2. **Write the Probability for Prime Values**:
We need to find \( P(X=2) + P(X=3) + P(X=5) \).
3. **Calculate Each Probability**:
Using the given probability function:
- \( P(X=2) = a \frac{(2+1)}{2^2} = a \frac{3}{4} \)
- \( P(X=3) = a \frac{(3+1)}{2^3} = a \frac{4}{8} = a \frac{1}{2} \)
- \( P(X=5) = a \frac{(5+1)}{2^5} = a \frac{6}{32} = a \frac{3}{16} \)
4. **Sum the Probabilities**:
Now, we sum these probabilities:
\[
P(X=2) + P(X=3) + P(X=5) = a \frac{3}{4} + a \frac{1}{2} + a \frac{3}{16}
\]
5. **Find a Common Denominator**:
The common denominator for \( 4, 2, \) and \( 16 \) is \( 16 \):
- \( P(X=2) = a \frac{3 \times 4}{16} = a \frac{12}{16} \)
- \( P(X=3) = a \frac{1 \times 8}{16} = a \frac{8}{16} \)
- \( P(X=5) = a \frac{3}{16} \)
Thus,
\[
P(X=2) + P(X=3) + P(X=5) = a \left( \frac{12 + 8 + 3}{16} \right) = a \frac{23}{16}
\]
6. **Determine the Value of \( a \)**:
To find \( a \), we use the fact that the total probability must equal 1:
\[
P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) = 1
\]
Calculate each term:
- \( P(X=0) = a \frac{1}{1} = a \)
- \( P(X=1) = a \frac{2}{2} = a \)
- \( P(X=2) = a \frac{3}{4} \)
- \( P(X=3) = a \frac{1}{2} \)
- \( P(X=4) = a \frac{5}{16} \)
- \( P(X=5) = a \frac{3}{16} \)
Therefore,
\[
a + a + a \frac{3}{4} + a \frac{1}{2} + a \frac{5}{16} + a \frac{3}{16} = 1
\]
Combine the terms:
\[
3a + a \left( \frac{12}{16} + \frac{8}{16} + \frac{5}{16} + \frac{3}{16} \right) = 1
\]
\[
3a + a \frac{28}{16} = 1
\]
\[
a \left( 3 + \frac{28}{16} \right) = 1
\]
\[
a \left( \frac{48 + 28}{16} \right) = 1
\]
\[
a \frac{76}{16} = 1 \implies a = \frac{16}{76} = \frac{4}{19}
\]
7. **Substitute \( a \) back into the Probability of Prime Values**:
Now substitute \( a \) into the sum of probabilities for prime values:
\[
P(X \text{ is prime}) = \frac{4}{19} \frac{23}{16} = \frac{4 \times 23}{19 \times 16} = \frac{92}{304} = \frac{23}{76}
\]
### Final Answer:
The probability that \( X \) takes a prime value is \( \frac{23}{76} \).
