A box contains 

5

 ﻿black, 

3

 ﻿red and 

4

 ﻿blue marbles. Two marbles are drawn without replacement.

What is the probability that the second and third marbles are red if it is known the first marble

is white?


The problem involves finding the probability that the second and third marbles are red, given that the first marble drawn is white. Let's break this down step by step.

### Key Points:
- The first marble is white, which **does not exist in the box** (there are only black, red, and blue marbles). This implies that the **first marble** being white is an external assumption and **does not affect** the composition of the remaining marbles in the box.
  
The box originally contains:
- 5 black marbles
- 3 red marbles
- 4 blue marbles

Thus, there are a total of:
\[
5 + 3 + 4 = 12 \text{ marbles.}
\]

Once we assume the first marble drawn is white (which does not exist in the box), we are effectively saying that the original set of marbles is unchanged, but we are only interested in the **second and third draws** from the 12 marbles.

### Step-by-Step Solution:

1. **Probability of drawing a red marble on the second draw:**
   Since no marbles have been removed (due to the assumption about the first being white), the probability of drawing a red marble on the second draw is:
   \[
   P(\text{red on second draw}) = \frac{3}{12} = \frac{1}{4}.
   \]

2. **Probability of drawing a red marble on the third draw:**
   After the second draw, there will be 11 marbles left. If the second marble was red, there will now be only 2 red marbles left. Hence, the probability of drawing a red marble on the third draw is:
   \[
   P(\text{red on third draw | red on second draw}) = \frac{2}{11}.
   \]

3. **Final Probability:**
   Since both events must occur (drawing a red marble on both the second and third draws), we multiply the probabilities of these independent events:
   \[
   P(\text{both second and third marbles are red}) = \frac{1}{4} \times \frac{2}{11} = \frac{2}{44} = \frac{1}{22}.
   \]

### Conclusion:
The probability that the second and third marbles are red, given that the first marble is white, is \( \frac{1}{22} \).

Would you like more details on any part of the solution?

Here are 5 related questions:
1. What is the probability that all three marbles drawn (without replacement) are red?
2. What is the probability of drawing one marble of each color (black, red, blue)?
3. How does the probability change if the marbles are drawn with replacement?
4. What is the probability that none of the three marbles drawn are black?
5. What is the probability that the second marble drawn is blue?

**Tip**: When dealing with probability, always identify whether events are independent or dependent, as this affects how probabilities are calculated!

A box contains 5 black, 3 red and 4 blue marbles. Two marbles are drawn without replacement. What is the probability that the second and third marbles are red if it is known the first marble is white?

Explore the probability of drawing two red marbles consecutively from a box containing 5 black, 3 red, and 4 blue marbles, given that the first marble drawn is white. This detailed solution covers key probability concepts and is suitable for students in Grades 9-12.

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