Question:
A bag contains 5 white and 3 black balls; another bag contains 4 white and 5 black balls. From any one of these bags, a single draw of two balls is made. Find the probability that one of them would be white and the other black.
A bag contains 5 white and 3 black balls; another bag contains 4 white and 5 black balls. From any one of these bags, a single draw of two balls is made. Find the probability that one of them would be white and the other black.
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Use total probability: average the “1 white, 1 black” probability from each bag, each weighted by 1/2.
Updated On: Jul 15, 2026
- 275/504
- 5/18
- 5/9
- None of these
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept.
Each bag is equally likely to be chosen, so \(P(\text{Bag 1})=P(\text{Bag 2})=\dfrac{1}{2}\). We need the probability of drawing one white and one black ball, averaged over both bags.
Step 2: Find the probability for Bag 1 (5 white, 3 black, 8 total).
\(P(\text{1W, 1B from Bag 1})=\dfrac{{}^5C_1\times{}^3C_1}{{}^8C_2}=\dfrac{5\times3}{28}=\dfrac{15}{28}\).
Step 3: Find the probability for Bag 2 (4 white, 5 black, 9 total).
\(P(\text{1W, 1B from Bag 2})=\dfrac{{}^4C_1\times{}^5C_1}{{}^9C_2}=\dfrac{4\times5}{36}=\dfrac{20}{36}=\dfrac{5}{9}\).
Step 4: Combine using total probability.
Required probability \(=\dfrac{1}{2}\times\dfrac{15}{28}+\dfrac{1}{2}\times\dfrac{5}{9}=\dfrac{1}{2}\left(\dfrac{15}{28}+\dfrac{5}{9}\right)\).
Step 5: Add the fractions.
LCM of 28 and 9 is 252: \(\dfrac{15}{28}=\dfrac{135}{252}\), \(\dfrac{5}{9}=\dfrac{140}{252}\). Sum \(=\dfrac{275}{252}\). Multiplying by \(\dfrac{1}{2}\): \(\dfrac{275}{504}\).
Step 6: Final Answer.
The required probability is \(\dfrac{275}{504}\), so option A is correct.
Each bag is equally likely to be chosen, so \(P(\text{Bag 1})=P(\text{Bag 2})=\dfrac{1}{2}\). We need the probability of drawing one white and one black ball, averaged over both bags.
Step 2: Find the probability for Bag 1 (5 white, 3 black, 8 total).
\(P(\text{1W, 1B from Bag 1})=\dfrac{{}^5C_1\times{}^3C_1}{{}^8C_2}=\dfrac{5\times3}{28}=\dfrac{15}{28}\).
Step 3: Find the probability for Bag 2 (4 white, 5 black, 9 total).
\(P(\text{1W, 1B from Bag 2})=\dfrac{{}^4C_1\times{}^5C_1}{{}^9C_2}=\dfrac{4\times5}{36}=\dfrac{20}{36}=\dfrac{5}{9}\).
Step 4: Combine using total probability.
Required probability \(=\dfrac{1}{2}\times\dfrac{15}{28}+\dfrac{1}{2}\times\dfrac{5}{9}=\dfrac{1}{2}\left(\dfrac{15}{28}+\dfrac{5}{9}\right)\).
Step 5: Add the fractions.
LCM of 28 and 9 is 252: \(\dfrac{15}{28}=\dfrac{135}{252}\), \(\dfrac{5}{9}=\dfrac{140}{252}\). Sum \(=\dfrac{275}{252}\). Multiplying by \(\dfrac{1}{2}\): \(\dfrac{275}{504}\).
Step 6: Final Answer.
The required probability is \(\dfrac{275}{504}\), so option A is correct.
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