To solve this problem, we need to find the probability of two specific events happening in sequence without replacement. We'll break it down step-by-step.
### Given:
- 3 pink balls
- 6 blue balls
- 3 red balls
Total number of balls initially:
[tex]\[ 3 + 6 + 3 = 12 \][/tex]
### Step 1: Probability of Event A (drawing a red ball on the first draw)
[tex]\[ P(A) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{3}{12} = \frac{1}{4} \][/tex]
### Step 2: Probability of Event B given that Event A has occurred (drawing a pink ball after drawing a red ball)
If a red ball has been drawn first, we are now left with:
- 2 red balls
- 6 blue balls
- 3 pink balls
So, the total number of remaining balls is:
[tex]\[ 12 - 1 = 11 \][/tex]
The probability of drawing a pink ball after already having drawn a red ball:
[tex]\[ P(B | A) = \frac{\text{Number of pink balls}}{\text{Total remaining balls}} = \frac{3}{11} \][/tex]
### Step 3: Probability of both events A and B happening (P(A and B))
[tex]\[ P(A \text{ and } B) = P(A) \times P(B | A) \][/tex]
Substitute the probabilities:
[tex]\[ P(A) = \frac{1}{4} \][/tex]
[tex]\[ P(B | A) = \frac{3}{11} \][/tex]
So,
[tex]\[ P(A \text{ and } B) = \frac{1}{4} \times \frac{3}{11} = \frac{3}{44} \][/tex]
Hence, the probability of drawing a red ball first and then a pink ball is:
[tex]\[ \frac{3}{44} \][/tex]
Therefore, the answer is:
[tex]\[ \boxed{\frac{3}{44}} \][/tex]
