Question 1166412: Three bags identical in appearance contain respectively 1 white and 3 red balls ; 2 white and 1 red balls ; 4 white and 3 red balls. One bag is selected randomly and two balls are taken from it. Find the probability that (1) One ball is white and another ball is red (2) A white ball is selected from bag, I, II, III and second ball is red.
Answer by CPhill(2030)   (Show Source):
You can put this solution on YOUR website! This is a problem involving **conditional probability** and the **Law of Total Probability**. Since the bags are identical in appearance, the probability of selecting any one bag is $\frac{1}{3}$.
Let $B_I$, $B_{II}$, and $B_{III}$ be the events that Bag I, Bag II, and Bag III are selected, respectively.
$$P(B_I) = P(B_{II}) = P(B_{III}) = \frac{1}{3}$$
---
## 🎒 Bag Contents
| Bag | White Balls ($W$) | Red Balls ($R$) | Total Balls ($T$) |
| :---: | :---: | :---: | :---: |
| **I** ($B_I$) | 1 | 3 | 4 |
| **II** ($B_{II}$) | 2 | 1 | 3 |
| **III** ($B_{III}$) | 4 | 3 | 7 |
Let $E$ be the event that "One ball is **White** and the other is **Red**" (i.e., one $W$ and one $R$) when two balls are drawn without replacement.
The probability of drawing one white and one red ball from a bag is given by:
$$P(E|B_i) = \frac{\text{(Ways to choose 1 W)} \times \text{(Ways to choose 1 R)}}{\text{Ways to choose 2 balls from Total}} = \frac{\binom{W}{1} \times \binom{R}{1}}{\binom{T}{2}}$$
### 1. Calculate Conditional Probabilities $P(E|B_i)$
* **From Bag I ($B_I$):**
$$P(E|B_I) = \frac{\binom{1}{1} \times \binom{3}{1}}{\binom{4}{2}} = \frac{1 \times 3}{6} = \frac{3}{6} = \frac{1}{2}$$
* **From Bag II ($B_{II}$):**
$$P(E|B_{II}) = \frac{\binom{2}{1} \times \binom{1}{1}}{\binom{3}{2}} = \frac{2 \times 1}{3} = \frac{2}{3}$$
* **From Bag III ($B_{III}$):**
$$P(E|B_{III}) = \frac{\binom{4}{1} \times \binom{3}{1}}{\binom{7}{2}} = \frac{4 \times 3}{21} = \frac{12}{21} = \frac{4}{7}$$
---
## (1) Probability that One Ball is White and Another Ball is Red
We use the **Law of Total Probability**:
$$P(E) = P(E|B_I)P(B_I) + P(E|B_{II})P(B_{II}) + P(E|B_{III})P(B_{III})$$
$$P(E) = \left(\frac{1}{2} \times \frac{1}{3}\right) + \left(\frac{2}{3} \times \frac{1}{3}\right) + \left(\frac{4}{7} \times \frac{1}{3}\right)$$
$$P(E) = \frac{1}{3} \left(\frac{1}{2} + \frac{2}{3} + \frac{4}{7}\right)$$
Find a common denominator for $\frac{1}{2} + \frac{2}{3} + \frac{4}{7}$. The Least Common Multiple (LCM) of 2, 3, and 7 is 42.
$$P(E) = \frac{1}{3} \left(\frac{21}{42} + \frac{28}{42} + \frac{24}{42}\right)$$
$$P(E) = \frac{1}{3} \left(\frac{21 + 28 + 24}{42}\right)$$
$$P(E) = \frac{1}{3} \times \frac{73}{42} = \frac{73}{126}$$
The probability that one ball is white and the other is red is $\mathbf{\frac{73}{126}}$.
---
## (2) A white ball is selected from bag, I, II, III and second ball is red
This part asks for the probability of the **compound event** that a specific bag is chosen **AND** the draw results in one white and one red ball.
Let $E$ be the event that "One ball is White and one ball is Red."
The prompt requires the sum of the probabilities of the three mutually exclusive events: $(E \text{ and } B_I)$, $(E \text{ and } B_{II})$, and $(E \text{ and } B_{III})$.
$$P(\text{Compound Events}) = P(E \cap B_I) + P(E \cap B_{II}) + P(E \cap B_{III})$$
Using the multiplication rule, $P(E \cap B_i) = P(E|B_i) \cdot P(B_i)$:
$$P(\text{Sum}) = \left(\frac{1}{2} \times \frac{1}{3}\right) + \left(\frac{2}{3} \times \frac{1}{3}\right) + \left(\frac{4}{7} \times \frac{1}{3}\right)$$
This is exactly the same calculation as Part (1), which is the total probability $P(E)$.
$$P(\text{Sum}) = \frac{1}{6} + \frac{2}{9} + \frac{4}{21}$$
$$P(\text{Sum}) = \frac{73}{126}$$
The probability that the draw results in one white and one red ball, summing across all three bags, is $\mathbf{\frac{73}{126}}$.
|
|
|
