What's the probability of drawing a red marble from the same bowl, given that the first marble drawn was blue and was not placed back in the bowl?

Answer

To solve the problem of finding the probability of drawing a red marble from a bowl after a blue marble has been drawn and not replaced, we can follow these steps:

### Step 1: Understand the initial conditions
Assume we have a bowl containing:
- 4 red marbles
- 5 blue marbles
- 4 green marbles

### Step 2: Calculate the total number of marbles initially
Initially, the total number of marbles in the bowl is:
\[ \text{Total marbles} = \text{Red} + \text{Blue} + \text{Green} = 4 + 5 + 4 = 13 \]

### Step 3: Draw a blue marble and adjust the counts
When we draw a blue marble and do not replace it, the counts change:
- The number of blue marbles becomes \( 5 - 1 = 4 \)
- The number of red marbles remains 4
- The number of green marbles remains 4

### Step 4: Calculate the new total number of marbles
After removing one blue marble, the new total number of marbles in the bowl is:
\[ \text{New total marbles} = \text{Red} + \text{Blue} + \text{Green} = 4 + 4 + 4 = 12 \]

### Step 5: Identify the number of favorable outcomes
The number of favorable outcomes for drawing a red marble is still 4 (since the number of red marbles has not changed).

### Step 6: Calculate the probability of drawing a red marble
The probability \( P \) of drawing a red marble now is given by the formula:
\[ P(\text{Red}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} \]
Substituting the values we have:
\[ P(\text{Red}) = \frac{4}{12} \]

### Step 7: Simplify the probability
We can simplify \( \frac{4}{12} \):
\[ P(\text{Red}) = \frac{1}{3} \]

### Final Answer
The probability of drawing a red marble from the bowl, given that the first marble drawn was blue and not replaced, is:
\[ \frac{1}{3} \]

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