Trevion has a bag of marbles with 1 blue marbles, 1 white marbles, and 2 red marbles.

Find the following probabilities of Trevion drawing the given marbles from the bag if the first marble(s) is(are) returned to the bag after they are drawn.

a) A Blue, then a red =

b) A red, then a white =

c) A Blue, then a Blue, then a Blue =

We will calculate each probability step by step based on the number of marbles in Trevion's bag and the rule that the first marble is returned to the bag after each draw (which means each draw is independent).

Trevion has:

- 1 blue marble
- 1 white marble
- 2 red marbles

This gives a total of \( 1 + 1 + 2 = 4 \) marbles.

### a) Probability of drawing a blue marble, then a red marble:

The probability of drawing a blue marble on the first draw is:
\[
P(\text{Blue}) = \frac{1}{4}
\]

The probability of drawing a red marble on the second draw (after the first marble is returned) is:
\[
P(\text{Red}) = \frac{2}{4} = \frac{1}{2}
\]

Since the events are independent, the total probability is:
\[
P(\text{Blue, then Red}) = P(\text{Blue}) \times P(\text{Red}) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}
\]

### b) Probability of drawing a red marble, then a white marble:

The probability of drawing a red marble on the first draw is:
\[
P(\text{Red}) = \frac{2}{4} = \frac{1}{2}
\]

The probability of drawing a white marble on the second draw is:
\[
P(\text{White}) = \frac{1}{4}
\]

Again, since the events are independent, the total probability is:
\[
P(\text{Red, then White}) = P(\text{Red}) \times P(\text{White}) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}
\]

### c) Probability of drawing a blue marble three times in a row:

Since the marble is returned after each draw, the probability of drawing a blue marble is the same for each draw:
\[
P(\text{Blue}) = \frac{1}{4}
\]

Thus, the total probability of drawing a blue marble three times is:
\[
P(\text{Blue, Blue, Blue}) = P(\text{Blue}) \times P(\text{Blue}) \times P(\text{Blue}) = \left(\frac{1}{4}\right)^3 = \frac{1}{64}
\]

### Final answers:

a) \( P(\text{Blue, then Red}) = \frac{1}{8} \)

b) \( P(\text{Red, then White}) = \frac{1}{8} \)

c) \( P(\text{Blue, Blue, Blue}) = \frac{1}{64} \)

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Would you like more details or have any questions about these calculations?

Here are 5 related questions:

1. What is the probability of drawing two white marbles in a row?
2. What is the probability of drawing a red marble first, then a blue marble?
3. What is the probability of drawing a red marble three times in a row?
4. If Trevion adds one more red marble, how would the probabilities change?
5. What is the probability of drawing one marble of each color in three draws?

**Tip:** When events are independent, you can find the total probability by multiplying the probabilities of each event.

Trevion has a bag of marbles with 1 blue marble, 1 white marble, and 2 red marbles. Find the probabilities of Trevion drawing the given marbles from the bag if the first marble(s) is(are) returned to the bag after they are drawn: a) A Blue, then a Red b) A Red, then a White c) A Blue, then a Blue, then a Blue

Explore how to calculate the probabilities of drawing different colored marbles (blue, red, white) from a bag in independent events. Learn step-by-step solutions for finding the probability of drawing specific combinations of marbles, such as Blue, then Red, and Blue three times in a row. Suitable for students in Grades 6-8.

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