There are 6 red and 8 green balls in a bag. 5 balls are drawn at random and placed in a red box. The remaining balls are placed in a green box. What is the probability that the number of red balls in the green box plus the number of green balls in the red box is not a prime number?

Answer

To solve the problem step by step, we will first analyze the situation and then calculate the required probability.

### Step 1: Understand the problem
We have a total of 6 red balls and 8 green balls. We draw 5 balls at random and place them in a red box. The remaining balls go into a green box. We need to find the probability that the sum of the number of red balls in the green box and the number of green balls in the red box is not a prime number.

### Step 2: Define variables
Let:
- \( r \) = number of red balls in the red box
- \( g \) = number of green balls in the red box

Since we draw 5 balls, we have:
- \( r + g = 5 \)

The remaining balls in the green box will be:
- Red balls in the green box = \( 6 - r \)
- Green balls in the green box = \( 8 - g \)

### Step 3: Calculate the sum
We need to find the sum:
\[ (6 - r) + g = 6 - r + g = 6 - r + (5 - r) = 11 - 2r \]
This means we need to check for values of \( r \) from 0 to 5 (since we can draw 0 to 5 red balls).

### Step 4: Determine the values of \( 11 - 2r \)
Now we calculate \( 11 - 2r \) for \( r = 0, 1, 2, 3, 4, 5 \):
- If \( r = 0 \): \( 11 - 2(0) = 11 \) (prime)
- If \( r = 1 \): \( 11 - 2(1) = 9 \) (not prime)
- If \( r = 2 \): \( 11 - 2(2) = 7 \) (prime)
- If \( r = 3 \): \( 11 - 2(3) = 5 \) (prime)
- If \( r = 4 \): \( 11 - 2(4) = 3 \) (prime)
- If \( r = 5 \): \( 11 - 2(5) = 1 \) (not prime)

### Step 5: Count the cases
From the calculations, the values of \( r \) that yield a non-prime sum \( (11 - 2r) \) are:
- \( r = 1 \) (sum = 9)
- \( r = 5 \) (sum = 1)

Thus, there are 2 favorable cases out of 6 possible values for \( r \).

### Step 6: Calculate the total number of ways to choose the balls
The total number of ways to choose 5 balls from 14 (6 red + 8 green) is given by:
\[ \binom{14}{5} \]

### Step 7: Calculate the probability
The probability \( P \) that the sum is not a prime number is:
\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{2}{6} = \frac{1}{3} \]

### Final Answer
Thus, the probability that the number of red balls in the green box plus the number of green balls in the red box is not a prime number is:
\[ \boxed{\frac{1}{3}} \]