SOLUTION: Urn A containes 5 red and 3 white marbles and urn B contain 1 red and 2 white marbles a fair die is tossed if a 3 or 6 appears, the marble is drawn from urn B and put into urn A an

Question 1166229: Urn A containes 5 red and 3 white marbles and urn B contain 1 red and 2 white marbles a fair die is tossed if a 3 or 6 appears, the marble is drawn from urn B and put into urn A and then a marble is drawn from urn A otherwise a marble is drawn from urn A and put into urn B and a marble is drawn from urn B what is the probability that
(1)both marbles are red
(2) both marbles are white
Answer by CPhill(2189) (Show Source): You can put this solution on YOUR website!
This is a probability problem involving two stages of drawing marbles, where the movement of the first marble depends on the outcome of a fair die roll.
Let's define the events and initial probabilities:
## 🎲 Initial Probabilities and Urn Contents
### Urn Contents
* **Urn A:** $5R, 3W \implies \text{Total } = 8$
* **Urn B:** $1R, 2W \implies \text{Total } = 3$
### Die Roll Events
Let $D$ be the event of rolling a fair six-sided die.
* **Event $E_B$ (Draw from B first):** A 3 or 6 appears.
$$P(E_B) = P(3) + P(6) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \mathbf{\frac{1}{3}}$$
* **Event $E_A$ (Draw from A first):** Not a 3 or 6 (i.e., 1, 2, 4, or 5).
$$P(E_A) = 1 - P(E_B) = 1 - \frac{1}{3} = \mathbf{\frac{2}{3}}$$
---
## 1. Probability that Both Marbles are Red ($R_1$ and $R_2$)
Let $M_1$ be the first marble drawn (and moved) and $M_2$ be the second marble drawn. $R_1$ and $R_2$ denote drawing a red marble in the first and second draws, respectively.
We use the Law of Total Probability, splitting the calculation based on the die roll outcome ($E_B$ or $E_A$).
$$P(R_1 \cap R_2) = P(R_1 \cap R_2 | E_B) \cdot P(E_B) + P(R_1 \cap R_2 | E_A) \cdot P(E_A)$$
### Case 1: $E_B$ Occurs (Move B $\to$ A, then Draw A)
1. **Draw $M_1$ from Urn B (Red):** $P(R_1 | E_B) = \frac{1}{3}$
2. **Move $R_1$ to Urn A:** Urn A contents become $6R, 3W$ (Total 9).
3. **Draw $M_2$ from Urn A (Red):** $P(R_2 | R_1 \text{ and } E_B) = \frac{6}{9} = \frac{2}{3}$
$$P(R_1 \cap R_2 | E_B) = P(R_1 | E_B) \cdot P(R_2 | R_1 \text{ and } E_B) = \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{9}$$
### Case 2: $E_A$ Occurs (Move A $\to$ B, then Draw B)
1. **Draw $M_1$ from Urn A (Red):** $P(R_1 | E_A) = \frac{5}{8}$
2. **Move $R_1$ to Urn B:** Urn B contents become $2R, 2W$ (Total 4).
3. **Draw $M_2$ from Urn B (Red):** $P(R_2 | R_1 \text{ and } E_A) = \frac{2}{4} = \frac{1}{2}$
$$P(R_1 \cap R_2 | E_A) = P(R_1 | E_A) \cdot P(R_2 | R_1 \text{ and } E_A) = \frac{5}{8} \cdot \frac{1}{2} = \frac{5}{16}$$
### Combine Cases
$$P(R_1 \cap R_2) = \left(\frac{2}{9} \cdot \frac{1}{3}\right) + \left(\frac{5}{16} \cdot \frac{2}{3}\right)$$
$$P(R_1 \cap R_2) = \frac{2}{27} + \frac{10}{48} = \frac{2}{27} + \frac{5}{24}$$
To add these, use the common denominator $LCM(27, 24) = 216$:
$$P(R_1 \cap R_2) = \frac{2 \cdot 8}{216} + \frac{5 \cdot 9}{216} = \frac{16}{216} + \frac{45}{216} = \frac{61}{216}$$
The probability that both marbles are red is **$\frac{61}{216}$**.
---
## 2. Probability that Both Marbles are White ($W_1$ and $W_2$)
Let $W_1$ and $W_2$ denote drawing a white marble in the first and second draws, respectively.
$$P(W_1 \cap W_2) = P(W_1 \cap W_2 | E_B) \cdot P(E_B) + P(W_1 \cap W_2 | E_A) \cdot P(E_A)$$
### Case 1: $E_B$ Occurs (Move B $\to$ A, then Draw A)
1. **Draw $M_1$ from Urn B (White):** $P(W_1 | E_B) = \frac{2}{3}$
2. **Move $W_1$ to Urn A:** Urn A contents become $5R, 4W$ (Total 9).
3. **Draw $M_2$ from Urn A (White):** $P(W_2 | W_1 \text{ and } E_B) = \frac{4}{9}$
$$P(W_1 \cap W_2 | E_B) = P(W_1 | E_B) \cdot P(W_2 | W_1 \text{ and } E_B) = \frac{2}{3} \cdot \frac{4}{9} = \frac{8}{27}$$
### Case 2: $E_A$ Occurs (Move A $\to$ B, then Draw B)
1. **Draw $M_1$ from Urn A (White):** $P(W_1 | E_A) = \frac{3}{8}$
2. **Move $W_1$ to Urn B:** Urn B contents become $1R, 3W$ (Total 4).
3. **Draw $M_2$ from Urn B (White):** $P(W_2 | W_1 \text{ and } E_A) = \frac{3}{4}$
$$P(W_1 \cap W_2 | E_A) = P(W_1 | E_A) \cdot P(W_2 | W_1 \text{ and } E_A) = \frac{3}{8} \cdot \frac{3}{4} = \frac{9}{32}$$
### Combine Cases
$$P(W_1 \cap W_2) = \left(\frac{8}{27} \cdot \frac{1}{3}\right) + \left(\frac{9}{32} \cdot \frac{2}{3}\right)$$
$$P(W_1 \cap W_2) = \frac{8}{81} + \frac{18}{96} = \frac{8}{81} + \frac{3}{16}$$
To add these, use the common denominator $LCM(81, 16) = 1296$:
$$P(W_1 \cap W_2) = \frac{8 \cdot 16}{1296} + \frac{3 \cdot 81}{1296} = \frac{128}{1296} + \frac{243}{1296} = \frac{371}{1296}$$
The probability that both marbles are white is **$\frac{371}{1296}$**.
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