Question

A pot contains 5 red and 2 green balls. A ball is drawn at random from this pot. If a drawn ball is green, then a red ball is added to the pot. If a drawn ball is red, then a green ball is added to the pot, while the original ball drawn is not replaced in the pot. Now a second ball is drawn at random from the pot, what is the probability that the second ball drawn is a red ball?

• A. \( \frac{12}{49} \)
• B. \( \frac{32}{49} \)
• C. \( \frac{3}{7} \)
• D. \( \frac{27}{49} \)

Hint:

In conditional probability problems, always split into cases and apply total probability lawCarefully track how composition changes after each event.

Answer 1

The Correct Option is B

Step 1: Initial probabilities.
Total balls = 5 red + 2 green = 7
\[ P(\text{first is red})=\frac{5}{7},\quad P(\text{first is green})=\frac{2}{7} \]

Step 2: Case 1 (first ball is red).
Red ball removed, green ball added.
New composition:
\[ 4 \text{ red},\; 3 \text{ green} \Rightarrow 7 \text{ total} \]
\[ P(\text{second is red | first red})=\frac{4}{7} \]

Step 3: Case 2 (first ball is green).
Green ball removed, red ball added.
New composition:
\[ 6 \text{ red},\; 1 \text{ green} \Rightarrow 7 \text{ total} \]
\[ P(\text{second is red | first green})=\frac{6}{7} \]

Step 4: Apply total probability theorem.
\[ P(\text{second red}) = P(R_1)\cdot P(R_2|R_1) + P(G_1)\cdot P(R_2|G_1) \]

Step 5: Substitute values.
\[ = \frac{5}{7}\cdot \frac{4}{7} + \frac{2}{7}\cdot \frac{6}{7} \]

Step 6: Simplify.
\[ = \frac{20}{49} + \frac{12}{49} \]
\[ = \frac{32}{49} \]

Step 7: Final conclusion.
\[ \boxed{\frac{32}{49}} \]