Question:
A bag contains 5 red and 4 black balls. Three balls are drawn at random from the bag. The probability that two of them are red and one is black is:
A bag contains 5 red and 4 black balls. Three balls are drawn at random from the bag. The probability that two of them are red and one is black is:
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Always simplify combinations first: $^{5}C_2 = 10$ and $^{9}C_3 = 84$. This makes dividing $\frac{40}{84}$ to get $\frac{10}{21}$ straightforward.
Updated On: May 31, 2026
- $\frac{5}{21}$
- $\frac{10}{21}$
- $\frac{5}{14}$
- $\frac{25}{84}$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
The classical definition of probability of an event is $P(E) = \frac{n(E)}{n(S)}$, where $n(E)$ is the number of favorable outcomes and $n(S)$ is the total number of possible outcomes.
Step 2: Meaning
The total number of balls in the bag is $5 + 4 = 9$. We are choosing $3$ balls. We want exactly $2$ red balls (from $5$) and $1$ black ball (from $4$).
Step 3: Analysis
Total number of ways to choose 3 balls out of 9: \[ n(S) = ^{9}C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \] Number of ways to choose 2 red balls and 1 black ball: \[ n(E) = ^{5}C_2 \times ^{4}C_1 = 10 \times 4 = 40 \] Calculating the probability: \[ P(E) = \frac{n(E)}{n(S)} = \frac{40}{84} = \frac{10}{21} \]
Step 4: Conclusion
The probability of drawing 2 red balls and 1 black ball is $\frac{10}{21}$. Final Answer: (B)
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