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Assume a jar has five red marbles and three black marbles. Draw out two marbles with and without replacement. Find the requested probabilities. (Enter the probabilities as fractions.)
(a)    P(two red marbles)
with replacement    	
without replacement    	

(b)    P(two black marbles)
with replacement    	
without replacement    	

(c)    P(one red and one black marble)
with replacement    	
without replacement    	

(d)    P(red on the first draw and black on the second draw)
with replacement    	
without replacement    	

Let's start with solving the first part of the problem from the tree diagram provided in the image.

### Problem:
We need to find \( P(C_1 | B_2) \cdot P(B_2) \).

From the tree diagram:

1. \( P(B_2) = \frac{1}{4} \)
2. \( P(C_1 | B_2) = \frac{1}{4} \)

Now, calculate the required probability:

\[ P(C_1 | B_2) \cdot P(B_2) = \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16} \]

Therefore, the answer is:

\[ \boxed{\frac{1}{16}} \]

---

Next, we address the problem of finding probabilities involving drawing marbles from a jar with and without replacement.

#### Assumptions:
- Total marbles: 8 (5 red + 3 black)
- Probability of drawing one red marble \( P(R) = \frac{5}{8} \)
- Probability of drawing one black marble \( P(B) = \frac{3}{8} \)

### (a) P(two red marbles)

#### With replacement:
\[ P(\text{two red marbles}) = P(R) \times P(R) = \left(\frac{5}{8}\right) \times \left(\frac{5}{8}\right) = \frac{25}{64} \]

#### Without replacement:
\[ P(\text{two red marbles}) = P(R) \times P(R_{\text{without first}}) = \left(\frac{5}{8}\right) \times \left(\frac{4}{7}\right) = \frac{20}{56} = \frac{5}{14} \]

### (b) P(two black marbles)

#### With replacement:
\[ P(\text{two black marbles}) = P(B) \times P(B) = \left(\frac{3}{8}\right) \times \left(\frac{3}{8}\right) = \frac{9}{64} \]

#### Without replacement:
\[ P(\text{two black marbles}) = P(B) \times P(B_{\text{without first}}) = \left(\frac{3}{8}\right) \times \left(\frac{2}{7}\right) = \frac{6}{56} = \frac{3}{28} \]

### (c) P(one red and one black marble)

#### With replacement:
\[ P(R \text{ and } B) = P(R) \times P(B) = \left(\frac{5}{8}\right) \times \left(\frac{3}{8}\right) = \frac{15}{64} \]
\[ P(B \text{ and } R) = P(B) \times P(R) = \left(\frac{3}{8}\right) \times \left(\frac{5}{8}\right) = \frac{15}{64} \]
\[ P(\text{one red and one black}) = \frac{15}{64} + \frac{15}{64} = \frac{30}{64} = \frac{15}{32} \]

#### Without replacement:
\[ P(R \text{ and } B) = P(R) \times P(B_{\text{without first}}) = \left(\frac{5}{8}\right) \times \left(\frac{3}{7}\right) = \frac{15}{56} \]
\[ P(B \text{ and } R) = P(B) \times P(R_{\text{without first}}) = \left(\frac{3}{8}\right) \times \left(\frac{5}{7}\right) = \frac{15}{56} \]
\[ P(\text{one red and one black}) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28} \]

### (d) P(red on the first draw and black on the second draw)

#### With replacement:
\[ P(R \text{ first and } B \text{ second}) = P(R) \times P(B) = \left(\frac{5}{8}\right) \times \left(\frac{3}{8}\right) = \frac{15}{64} \]

#### Without replacement:
\[ P(R \text{ first and } B \text{ second}) = P(R) \times P(B_{\text{without first}}) = \left(\frac{5}{8}\right) \times \left(\frac{3}{7}\right) = \frac{15}{56} \]

Would you like further details or have any other questions? Here are some related questions you might consider:

1. What is the probability of drawing two marbles of the same color?
2. How does the probability change if the number of marbles changes?
3. What is the probability of drawing at least one red marble in two draws?
4. How would the probabilities change if there are more than two colors of marbles?
5. What is the expected value of drawing a red marble in two draws?

### Tip:
Always keep track of how replacement affects the total number of items when calculating probabilities.

Explore the probabilities of drawing marbles from a jar containing five red marbles and three black marbles, with and without replacement. Calculate the probabilities of drawing two red marbles, two black marbles, and one red and one black marble under both conditions.

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