Question

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?

• A. $\frac{35}{57}$
• B. $\frac{99}{257}$
• C. $\frac{3}{7}$
• D. $\frac{33}{53}$

Hint:

In Bayes' Theorem problems, keeping fractions with common denominators (like 385 here) makes final division extremely straightforward as the denominators simply cancel out.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a conditional probability problem that can be solved using Bayes' Theorem.
Key Formula or Approach:
Let \( R_{1} \) be the event that red ball is drawn from \( B_{1} \), and \( R_{2} \) be the event that red ball is drawn from \( B_{2} \).
\[ P(R_{1}|R_{2}) = \frac{P(R_{1})P(R_{2}|R_{1})}{P(R_{1})P(R_{2}|R_{1}) + P(B_{1})P(R_{2}|B_{1})} \]

Step 2: Detailed Explanation:
1. Probabilities from \( B_{1} \):
\( P(R_{1}) = 3/7 \) and \( P(B_{1}) = 4/7 \).
2. Conditional probabilities for \( B_{2} \):
$\bullet$ If \( R_{1} \) occurs, 4 red balls are added to \( B_{2} \). Total balls in \( B_{2} \) become \( 7+4=11 \), with \( 3+4=7 \) red balls.
\( P(R_{2}|R_{1}) = 7/11 \).
$\bullet$ If \( B_{1} \) occurs (black ball), 3 black balls are added to \( B_{2} \). Total balls in \( B_{2} \) become \( 7+3=10 \), with 3 red balls remaining.
\( P(R_{2}|B_{1}) = 3/10 \).
3. Apply Bayes' Theorem:
\[ P(R_{1}|R_{2}) = \frac{(3/7) \times (7/11)}{(3/7 \times 7/11) + (4/7 \times 3/10)} \]
\[ P(R_{1}|R_{2}) = \frac{3/11}{3/11 + 12/70} = \frac{3/11}{3/11 + 6/35} \]
\[ P(R_{1}|R_{2}) = \frac{3/11}{(105+66)/385} = \frac{3/11}{171/385} = \frac{3 \times 35}{171} = \frac{105}{171} = \frac{35}{57} \]

Step 3: Final Answer:
The conditional probability is \( \frac{35}{57} \).