Suppose there are two boxes $B_1$ and $B_2$, each having 3 red and 4 black balls. One ball is drawn at random from $B_1$. If it is red, 4 red balls are put into $B_2$, otherwise 3 black balls are put into $B_2$. Then one ball is randomly drawn from $B_2$. If this ball is red, what is the conditional probability that the ball drawn from $B_1$ was also red?
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In Bayes' Theorem problems, keeping fractions with common denominators (like 385 here) makes final division extremely straightforward as the denominators simply cancel out.
Step 1: Understanding the Question:
We are given two boxes, $B_1$ and $B_2$, each initially containing 3 red and 4 black balls.
A ball is drawn from $B_1$, and depending on its color, a specific number of balls are added to $B_2$.
Finally, a ball is drawn from $B_2$, and it is found to be red.
We need to calculate the conditional probability that the ball drawn from $B_1$ was red. Step 2: Key Formula or Approach:
Let us define the events:
- $R_1$: The ball drawn from $B_1$ is red.
- $B_1$: The ball drawn from $B_1$ is black.
- $R_2$: The ball drawn from $B_2$ is red.
We need to find the conditional probability $P(R_1 \mid R_2)$. By Bayes' Theorem:
\[ P(R_1 \mid R_2) = \frac{P(R_1) P(R_2 \mid R_1)}{P(R_1) P(R_2 \mid R_1) + P(B_1) P(R_2 \mid B_1)} \]
Step 3: Detailed Explanation:
• First, we calculate the prior probabilities for the first draw from $B_1$:
The box $B_1$ has 3 red and 4 black balls (total of 7 balls).
\[ P(R_1) = \frac{3}{7} \]
\[ P(B_1) = \frac{4}{7} \]
• Now, we calculate the conditional probabilities for the second draw:
- Case 1: If a red ball was drawn from $B_1$ (event $R_1$):
We add 4 red balls to $B_2$.
The new composition of $B_2$ becomes: $3 + 4 = 7$ red balls, and 4 black balls (total of 11 balls).
\[ P(R_2 \mid R_1) = \frac{7}{11} \]
- Case 2: If a black ball was drawn from $B_1$ (event $B_1$):
We add 3 black balls to $B_2$.
The new composition of $B_2$ becomes: 3 red balls, and $4 + 3 = 7$ black balls (total of 10 balls).
\[ P(R_2 \mid B_1) = \frac{3}{10} \]
• Now, we apply Bayes' Theorem:
\[ P(R_1 \mid R_2) = \frac{\frac{3}{7} \times \frac{7}{11}}{\left(\frac{3}{7} \times \frac{7}{11}\right) + \left(\frac{4}{7} \times \frac{3}{10}\right)} \]
Simplifying the terms:
\[ \text{Numerator} = \frac{3}{11} \]
\[ \text{Denominator} = \frac{3}{11} + \frac{12}{70} = \frac{3}{11} + \frac{6}{35} \]
Find a common denominator for the terms:
\[ \text{Denominator} = \frac{3 \times 35 + 6 \times 11}{11 \times 35} = \frac{105 + 66}{385} = \frac{171}{385} \]
\[ \text{Numerator} = \frac{3}{11} = \frac{3 \times 35}{385} = \frac{105}{385} \]
Thus:
\[ P(R_1 \mid R_2) = \frac{105}{171} \]
Dividing both the numerator and the denominator by 3:
\[ P(R_1 \mid R_2) = \frac{35}{57} \]
Step 4: Final Answer:
The conditional probability is $\frac{35}{57}$.
