There are 3 boxes each containing 3 red and 5 green balls. Also there are 2 boxes, each containing 4 red and 2 green balls. A green ball is selected at random. Find the probability that this green ball is from a box of the first group.

Options

A. A) `54/61`
B. B) `45/61`
C. C) `8/31`
D. D) none of these

Answer

To solve the problem step by step, we will use the concept of conditional probability and the law of total probability.

### Step 1: Define Events
Let:
- \( E_1 \): The event of selecting a box from the first group (3 boxes with 3 red and 5 green balls).
- \( E_2 \): The event of selecting a box from the second group (2 boxes with 4 red and 2 green balls).
- \( A \): The event of selecting a green ball.

### Step 2: Calculate the Total Number of Boxes
The total number of boxes is:
- From the first group: 3 boxes
- From the second group: 2 boxes
- Total boxes = \( 3 + 2 = 5 \)

### Step 3: Calculate the Probability of Selecting Each Group of Boxes
The probability of selecting a box from the first group \( P(E_1) \) is:
\[ P(E_1) = \frac{3}{5} \]
The probability of selecting a box from the second group \( P(E_2) \) is:
\[ P(E_2) = \frac{2}{5} \]

### Step 4: Calculate the Probability of Selecting a Green Ball Given Each Group
Now we calculate the probability of selecting a green ball given that we have selected a box from either group.

For the first group \( P(A | E_1) \):
- Number of green balls = 5
- Total balls = 3 red + 5 green = 8
\[ P(A | E_1) = \frac{5}{8} \]

For the second group \( P(A | E_2) \):
- Number of green balls = 2
- Total balls = 4 red + 2 green = 6
\[ P(A | E_2) = \frac{2}{6} = \frac{1}{3} \]

### Step 5: Calculate the Total Probability of Selecting a Green Ball
Using the law of total probability, we calculate \( P(A) \):
\[ P(A) = P(A | E_1) \cdot P(E_1) + P(A | E_2) \cdot P(E_2) \]
Substituting the values:
\[ P(A) = \left(\frac{5}{8} \cdot \frac{3}{5}\right) + \left(\frac{1}{3} \cdot \frac{2}{5}\right) \]
Calculating each term:
\[ = \frac{15}{40} + \frac{2}{15} \]
To add these fractions, we need a common denominator, which is 120:
\[ = \frac{15 \cdot 3}{120} + \frac{2 \cdot 8}{120} = \frac{45}{120} + \frac{16}{120} = \frac{61}{120} \]

### Step 6: Calculate the Probability that the Green Ball is from the First Group
We want to find \( P(E_1 | A) \):
\[ P(E_1 | A) = \frac{P(A | E_1) \cdot P(E_1)}{P(A)} \]
Substituting the values:
\[ P(E_1 | A) = \frac{\left(\frac{5}{8}\right) \cdot \left(\frac{3}{5}\right)}{\frac{61}{120}} \]
Calculating the numerator:
\[ = \frac{15}{40} \]
Thus,
\[ P(E_1 | A) = \frac{15/40}{61/120} = \frac{15 \cdot 120}{40 \cdot 61} = \frac{1800}{2440} = \frac{45}{61} \]

### Final Answer
The probability that the green ball is from a box of the first group is:
\[ \frac{45}{61} \]