Round all answers to 4 decimal places.

a.  A bag contains 6 white marbles, 5 black marbles, and 4 yellow marbles. If a marble is drawn from the bag, replaced, and another marble is drawn, what is the probability of drawing first a white marble and then a yellow marble?
  

b.  A bag contains 5 blue marbles, 4 yellow marbles, and 6 black marbles. If two different marbles are drawn from the bag , what is the probability of drawing first a blue marble and then a black 
 marble?
  


Let's solve both probability questions step by step:

---

### **Part a**: Drawing first a white marble and then a yellow marble (with replacement)

The total number of marbles in the bag is:
\[
6 \, (\text{white}) + 5 \, (\text{black}) + 4 \, (\text{yellow}) = 15 \, \text{marbles}.
\]

**Step 1: Probability of drawing a white marble on the first draw:**

The probability of drawing a white marble is the ratio of white marbles to the total number of marbles:
\[
P(\text{White}) = \frac{6}{15} = 0.4000.
\]

**Step 2: Probability of drawing a yellow marble on the second draw (with replacement):**

Since the marble is replaced, the total number of marbles remains 15. The probability of drawing a yellow marble is:
\[
P(\text{Yellow}) = \frac{4}{15} = 0.2667.
\]

**Step 3: Combined probability (White then Yellow):**

Since both events are independent (because of replacement), we multiply the individual probabilities:
\[
P(\text{White then Yellow}) = P(\text{White}) \times P(\text{Yellow}) = \frac{6}{15} \times \frac{4}{15} = 0.4000 \times 0.2667 = 0.1067.
\]

---

### **Part b**: Drawing first a blue marble and then a black marble (without replacement)

The total number of marbles in the bag is:
\[
5 \, (\text{blue}) + 4 \, (\text{yellow}) + 6 \, (\text{black}) = 15 \, \text{marbles}.
\]

**Step 1: Probability of drawing a blue marble on the first draw:**

The probability of drawing a blue marble is:
\[
P(\text{Blue}) = \frac{5}{15} = 0.3333.
\]

**Step 2: Probability of drawing a black marble on the second draw (without replacement):**

After drawing a blue marble, there are now 14 marbles left in the bag. The probability of drawing a black marble is:
\[
P(\text{Black}) = \frac{6}{14} = 0.4286.
\]

**Step 3: Combined probability (Blue then Black):**

Since the events are dependent (without replacement), we multiply the probabilities:
\[
P(\text{Blue then Black}) = P(\text{Blue}) \times P(\text{Black}) = \frac{5}{15} \times \frac{6}{14} = 0.3333 \times 0.4286 = 0.1429.
\]

---

### Final answers:
- **a. Probability of drawing first a white marble and then a yellow marble (with replacement):**  
  \[
  P(\text{White then Yellow}) = 0.1067.
  \]

- **b. Probability of drawing first a blue marble and then a black marble (without replacement):**  
  \[
  P(\text{Blue then Black}) = 0.1429.
  \]

---

Do you want more details or have any questions?

Here are 5 related questions:
1. What is the probability of drawing two white marbles in a row (with replacement)?
2. What if the marbles were drawn without replacement in part (a)? How would the probability change?
3. How does the probability of drawing two different-colored marbles compare to two marbles of the same color?
4. What if there were more marbles added to the bag? How

A bag contains 6 white marbles, 5 black marbles, and 4 yellow marbles. If a marble is drawn from the bag, replaced, and another marble is drawn, what is the probability of drawing first a white marble and then a yellow marble? A bag contains 5 blue marbles, 4 yellow marbles, and 6 black marbles. If two different marbles are drawn from the bag, what is the probability of drawing first a blue marble and then a black marble?

Learn how to calculate the probability of drawing first a white marble and then a yellow marble (with replacement), and first a blue marble and then a black marble (without replacement) from two different bags of marbles. This solution covers key concepts in independent and dependent events in probability.

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