Question

First bag contains 3 red and 5 black balls and second bag contains 6 red and 4 black balls. A ball is drawn from each bag. The probability that one ball is red and the other is black, is

• A. $\frac{41}{80}$
• B. $\frac{21}{40}$
• C. $\frac{3}{20}$
• D. $\frac{3}{8}$

Hint:

Always keep a common denominator when calculating intermediate case probabilities (use $\frac{6}{40}$ instead of simplifying to $\frac{3}{20}$). This allows you to add the fractions at the final step instantly without needing to find a new common base!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given two separate bags filled with combinations of red and black balls. One ball is drawn at random from each bag. We need to determine the total probability that the resulting pair consists of exactly one red ball and exactly one black ball.

Step 2: Key Formula or Approach:
Since the selections from each bag are independent, we use the multiplication rule for independent events. There are two mutually exclusive configurations that satisfy the requirement: Case 1: Red ball from Bag 1 AND Black ball from Bag 2. Case 2: Black ball from Bag 1 AND Red ball from Bag 2. The total probability will be the sum of these two case probabilities: $P(\text{Total}) = P(R_1) \cdot P(B_2) + P(B_1) \cdot P(R_2)$.

Step 3: Detailed Explanation:
Let's list the ball counts and find the individual probabilities for each bag: $$\text{Bag 1: 3 Red + 5 Black = 8 total balls} \implies P(R_1) = \frac{3}{8}, \quad P(B_1) = \frac{5}{8}$$ $$\text{Bag 2: 6 Red + 4 Black = 10 total balls} \implies P(R_2) = \frac{6}{10} = \frac{3}{5}, \quad P(B_2) = \frac{4}{10} = \frac{2}{5}$$ Now, let's calculate the values for our two cases: 1. Probability of Case 1 (Red from Bag 1, Black from Bag 2): $$P(\text{Case 1}) = P(R_1) \cdot P(B_2) = \frac{3}{8} \times \frac{2}{5} = \frac{6}{40}$$ 2. Probability of Case 2 (Black from Bag 1, Red from Bag 2): $$P(\text{Case 2}) = P(B_1) \cdot P(R_2) = \frac{5}{8} \times \frac{3}{5} = \frac{15}{40}$$ Sum the probabilities of these two independent cases to get the total probability: $$P(\text{Total}) = \frac{6}{40} + \frac{15}{40} = \frac{21}{40}$$

Step 4: Final Answer:
The probability that one ball is red and the other is black is $\frac{21}{40}$, which corresponds to option (B).