An urn contains three red balls and n white balls. Mr. A draws two balls together from the urn. The probability that they have the same color is `1//2.` Mr.B draws one ball from the urn, notes its color and rplaces it. He then draws a second ball from the urn and finds that both balls have the same color is `5//8.` The value of n is ____.

Explanation

To solve the problem, we need to find the value of \( n \) given the conditions for Mr. A and Mr. B's draws from the urn.

### Step 1: Set Up the Problem
We have:
- 3 red balls
- \( n \) white balls

The total number of balls is \( n + 3 \).

### Step 2: Probability for Mr. A
Mr. A draws two balls together. The probability that they have the same color is given as \( \frac{1}{2} \).

The total ways to choose 2 balls from \( n + 3 \) balls is:
\[ \binom{n+3}{2} = \frac{(n+3)(n+2)}{2} \]

The ways to choose 2 balls of the same color:
- Choosing 2 red balls: \( \binom{3}{2} = 3 \)
- Choosing 2 white balls: \( \binom{n}{2} = \frac{n(n-1)}{2} \)

Thus, the probability that both balls are of the same color is:
\[ P(A) = \frac{\binom{3}{2} + \binom{n}{2}}{\binom{n+3}{2}} = \frac{3 + \frac{n(n-1)}{2}}{\frac{(n+3)(n+2)}{2}} \]

Setting this equal to \( \frac{1}{2} \):
\[ \frac{3 + \frac{n(n-1)}{2}}{\frac{(n+3)(n+2)}{2}} = \frac{1}{2} \]

### Step 3: Simplify the Equation
Cross-multiplying gives:
\[ 2 \left( 3 + \frac{n(n-1)}{2} \right) = (n+3)(n+2) \]
\[ 6 + n(n-1) = n^2 + 5n + 6 \]
\[ n(n-1) = n^2 + 5n \]
\[ 0 = 4n \implies n(n - 7) = 0 \]
Thus, \( n = 0 \) or \( n = 7 \).

### Step 4: Probability for Mr. B
Now we consider Mr. B's scenario. He draws one ball, notes its color, replaces it, and then draws again. The probability that both balls have the same color is given as \( \frac{5}{8} \).

The probability that he draws two balls of the same color can be calculated as:
\[ P(B) = P(\text{both red}) + P(\text{both white}) \]
\[ P(B) = \left( \frac{3}{n+3} \cdot \frac{3}{n+3} \right) + \left( \frac{n}{n+3} \cdot \frac{n}{n+3} \right) \]
\[ = \frac{9}{(n+3)^2} + \frac{n^2}{(n+3)^2} \]
\[ = \frac{9 + n^2}{(n+3)^2} \]

Setting this equal to \( \frac{5}{8} \):
\[ \frac{9 + n^2}{(n+3)^2} = \frac{5}{8} \]

### Step 5: Cross-Multiply and Simplify
Cross-multiplying gives:
\[ 8(9 + n^2) = 5(n + 3)^2 \]
\[ 72 + 8n^2 = 5(n^2 + 6n + 9) \]
\[ 72 + 8n^2 = 5n^2 + 30n + 45 \]
\[ 3n^2 - 30n + 27 = 0 \]
Dividing by 3:
\[ n^2 - 10n + 9 = 0 \]

### Step 6: Solve the Quadratic Equation
Using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{10 \pm \sqrt{100 - 36}}{2} = \frac{10 \pm 8}{2} \]
Thus, \( n = 9 \) or \( n = 1 \).

### Step 7: Conclusion
From the two equations derived, we have \( n = 7 \) from Mr. A's probability and \( n = 1 \) or \( n = 9 \) from Mr. B's probability. The only consistent solution is \( n = 1 \).

Therefore, the value of \( n \) is:
\[ \boxed{1} \]