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 ____.

Options

A. 9
B. 6
C. 5
D. 1

Explanation

To solve the problem, we need to find the value of \( n \) given the conditions about the balls drawn from the urn.

### Step 1: Setting up the problem
We have an urn containing 3 red balls and \( n \) white balls. The total number of balls in the urn is \( n + 3 \).

### Step 2: Probability that Mr. A draws two balls of the same color
The probability that Mr. A draws two balls of the same color can be calculated by considering two cases: drawing two red balls or drawing two white balls.

1. **Probability of drawing 2 red balls**:
\[ P(\text{2 red}) = \frac{\binom{3}{2}}{\binom{n+3}{2}} = \frac{3}{\frac{(n+3)(n+2)}{2}} = \frac{6}{(n+3)(n+2)} \]

2. **Probability of drawing 2 white balls**:
\[ P(\text{2 white}) = \frac{\binom{n}{2}}{\binom{n+3}{2}} = \frac{\frac{n(n-1)}{2}}{\frac{(n+3)(n+2)}{2}} = \frac{n(n-1)}{(n+3)(n+2)} \]

3. **Total probability of drawing 2 balls of the same color**:
\[ P(\text{same color}) = P(\text{2 red}) + P(\text{2 white}) = \frac{6}{(n+3)(n+2)} + \frac{n(n-1)}{(n+3)(n+2)} \]
\[ = \frac{6 + n(n-1)}{(n+3)(n+2)} \]

According to the problem, this probability equals \( \frac{1}{2} \):
\[ \frac{6 + n(n-1)}{(n+3)(n+2)} = \frac{1}{2} \]

### Step 3: Cross-multiplying to solve for \( n \)
Cross-multiplying gives:
\[ 2(6 + n(n-1)) = (n+3)(n+2) \]
Expanding both sides:
\[ 12 + 2n^2 - 2n = n^2 + 5n + 6 \]
Rearranging gives:
\[ 2n^2 - n^2 - 2n - 5n + 12 - 6 = 0 \]
\[ n^2 - 7n + 6 = 0 \]

### Step 4: Factoring the quadratic equation
Factoring gives:
\[ (n - 1)(n - 6) = 0 \]
Thus, the possible values for \( n \) are:
\[ n = 1 \quad \text{or} \quad n = 6 \]

### Step 5: Analyzing Mr. B's draws
Now we need to check which value of \( n \) satisfies the second condition given for Mr. B.

1. **Probability that Mr. B draws two balls of the same color**:
- If Mr. B draws a red ball first, the probability of drawing a red ball again is:
\[ P(\text{2 red}) = \frac{3}{n+3} \cdot \frac{3}{n+3} = \frac{9}{(n+3)^2} \]
- If Mr. B draws a white ball first, the probability of drawing a white ball again is:
\[ P(\text{2 white}) = \frac{n}{n+3} \cdot \frac{n}{n+3} = \frac{n^2}{(n+3)^2} \]

2. **Total probability that both balls have the same color**:
\[ P(\text{same color}) = P(\text{2 red}) + P(\text{2 white}) = \frac{9 + n^2}{(n+3)^2} \]
According to the problem, this probability equals \( \frac{5}{8} \):
\[ \frac{9 + n^2}{(n+3)^2} = \frac{5}{8} \]

### Step 6: Cross-multiplying to solve for \( n \)
Cross-multiplying gives:
\[ 8(9 + n^2) = 5(n + 3)^2 \]
Expanding both sides:
\[ 72 + 8n^2 = 5(n^2 + 6n + 9) \]
\[ 72 + 8n^2 = 5n^2 + 30n + 45 \]
Rearranging gives:
\[ 3n^2 - 30n + 27 = 0 \]
Dividing by 3:
\[ n^2 - 10n + 9 = 0 \]

### Step 7: Factoring the quadratic equation
Factoring gives:
\[ (n - 1)(n - 9) = 0 \]
Thus, the possible values for \( n \) are:
\[ n = 1 \quad \text{or} \quad n = 9 \]

### Step 8: Conclusion
The common solution from both conditions is \( n = 1 \). Therefore, the value of \( n \) is:
\[ \boxed{1} \]