An urn contains 3 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 balls form the urn, notes its color and replaces it. He then draws a second ball from the urn and finds that both balls have the same color is 5/8. The possible value of `n` is `9` b. `6` c. `5` d. 1

Explanation

To solve the problem, we need to analyze the information given and set up equations based on the probabilities described.

### Step 1: Set up the problem for Mr. A
Mr. A draws two balls together from the urn. The total number of balls in the urn is \(3 + n\) (3 red balls and \(n\) white balls).

The probability that both balls drawn have the same color can be expressed as:
\[ P(\text{same color}) = \frac{\text{Number of ways to choose 2 red balls} + \text{Number of ways to choose 2 white balls}}{\text{Total ways to choose 2 balls}} \]
Calculating the number of ways:
- The number of ways to choose 2 red balls from 3 is \( \binom{3}{2} = 3 \).
- The number of ways to choose 2 white balls from \(n\) is \( \binom{n}{2} = \frac{n(n-1)}{2} \).
- The total ways to choose 2 balls from \(3 + n\) is \( \binom{3+n}{2} = \frac{(3+n)(2+n)}{2} \).

Thus, the probability can be written as:
\[ P(\text{same color}) = \frac{3 + \frac{n(n-1)}{2}}{\frac{(3+n)(2+n)}{2}} = \frac{6 + n(n-1)}{(3+n)(2+n)} \]
According to the problem, this probability is given as \( \frac{1}{2} \):
\[ \frac{6 + n(n-1)}{(3+n)(2+n)} = \frac{1}{2} \]

### Step 2: Cross-multiply and simplify
Cross-multiplying gives:
\[ 2(6 + n(n-1)) = (3+n)(2+n) \]
Expanding both sides:
\[ 12 + 2n(n-1) = 6 + 5n + n^2 \]
Rearranging the equation:
\[ 2n^2 - n^2 - 5n + 12 - 6 = 0 \]
This simplifies to:
\[ n^2 - 5n + 6 = 0 \]

### Step 3: Factor the quadratic equation
Factoring gives:
\[ (n - 2)(n - 3) = 0 \]
Thus, the possible values for \(n\) are:
\[ n = 2 \quad \text{or} \quad n = 3 \]

### Step 4: Set up the problem for Mr. B
Mr. B draws one ball, notes its color, replaces it, and then draws a second ball. The probability that both balls are of the same color is given as \( \frac{5}{8} \).

The probability can be expressed as:
\[ P(\text{same color}) = P(\text{both red}) + P(\text{both white}) \]
Calculating:
- The probability of drawing a red ball twice is \( \left(\frac{3}{3+n}\right)^2 \).
- The probability of drawing a white ball twice is \( \left(\frac{n}{3+n}\right)^2 \).

Thus, we have:
\[ P(\text{same color}) = \left(\frac{3}{3+n}\right)^2 + \left(\frac{n}{3+n}\right)^2 = \frac{9 + n^2}{(3+n)^2} \]
Setting this equal to \( \frac{5}{8} \):
\[ \frac{9 + n^2}{(3+n)^2} = \frac{5}{8} \]

### Step 5: Cross-multiply and simplify
Cross-multiplying gives:
\[ 8(9 + n^2) = 5(3+n)^2 \]
Expanding both sides:
\[ 72 + 8n^2 = 5(9 + 6n + n^2) \]
This simplifies to:
\[ 72 + 8n^2 = 45 + 30n + 5n^2 \]
Rearranging gives:
\[ 3n^2 - 30n + 27 = 0 \]

### Step 6: Factor the quadratic equation
Dividing through by 3:
\[ n^2 - 10n + 9 = 0 \]
Factoring gives:
\[ (n - 1)(n - 9) = 0 \]
Thus, the possible values for \(n\) are:
\[ n = 1 \quad \text{or} \quad n = 9 \]

### Step 7: Find common values
From the two equations, we have:
1. From Mr. A: \(n = 2\) or \(n = 3\)
2. From Mr. B: \(n = 1\) or \(n = 9\)

The only common value that satisfies both equations is \(n = 1\).

### Final Answer
The possible value of \(n\) is \(1\).

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