A bag contains 3 green balls and 7 white balls. Two balls are drawn at random without replacement. If the second ball drawn is green, what is the probability that the first ball drawn is also green ?

Answer

To solve the problem of finding the probability that the first ball drawn is green given that the second ball drawn is green, we can use Bayes' theorem. Let's denote the events as follows:

- Let \( E_1 \) be the event that the first ball drawn is green.
- Let \( E_2 \) be the event that the first ball drawn is white.
- Let \( A \) be the event that the second ball drawn is green.

We need to find \( P(E_1 | A) \), the probability that the first ball is green given that the second ball is green.

### Step 1: Calculate the total number of balls
The bag contains 3 green balls and 7 white balls, so the total number of balls is:
\[ 3 + 7 = 10 \]

### Step 2: Calculate \( P(E_1) \) and \( P(E_2) \)
- The probability that the first ball drawn is green:
\[ P(E_1) = \frac{3}{10} \]
- The probability that the first ball drawn is white:
\[ P(E_2) = \frac{7}{10} \]

### Step 3: Calculate \( P(A | E_1) \) and \( P(A | E_2) \)
- If the first ball drawn is green, there are 2 green balls left and 9 balls total remaining. Thus:
\[ P(A | E_1) = \frac{2}{9} \]
- If the first ball drawn is white, there are still 3 green balls left and 9 balls total remaining. Thus:
\[ P(A | E_2) = \frac{3}{9} = \frac{1}{3} \]

### Step 4: Use Bayes' theorem to find \( P(E_1 | A) \)
Bayes' theorem states:
\[ P(E_1 | A) = \frac{P(A | E_1) \cdot P(E_1)}{P(A)} \]
where \( P(A) \) can be calculated using the law of total probability:
\[ P(A) = P(A | E_1) \cdot P(E_1) + P(A | E_2) \cdot P(E_2) \]

Calculating \( P(A) \):
\[ P(A) = \left(\frac{2}{9} \cdot \frac{3}{10}\right) + \left(\frac{1}{3} \cdot \frac{7}{10}\right) \]
Calculating each term:
1. \( P(A | E_1) \cdot P(E_1) = \frac{2}{9} \cdot \frac{3}{10} = \frac{6}{90} = \frac{1}{15} \)
2. \( P(A | E_2) \cdot P(E_2) = \frac{1}{3} \cdot \frac{7}{10} = \frac{7}{30} \)

Now, convert \( \frac{7}{30} \) to a fraction with a denominator of 90:
\[ \frac{7}{30} = \frac{21}{90} \]

So,
\[ P(A) = \frac{1}{15} + \frac{21}{90} = \frac{6}{90} + \frac{21}{90} = \frac{27}{90} = \frac{3}{10} \]

### Step 5: Substitute back to find \( P(E_1 | A) \)
Now substituting back into Bayes' theorem:
\[ P(E_1 | A) = \frac{P(A | E_1) \cdot P(E_1)}{P(A)} = \frac{\left(\frac{2}{9}\right) \cdot \left(\frac{3}{10}\right)}{\frac{3}{10}} \]
This simplifies to:
\[ P(E_1 | A) = \frac{2}{9} \]

### Final Answer
Thus, the probability that the first ball drawn is green given that the second ball drawn is green is:
\[ \boxed{\frac{2}{9}} \]