Question

A bag contains 4 red and 5 blue balls. Two balls are drawn without replacement. What is the probability that both are red? 
 

• A. $\frac{2}{9}$
• B. $\frac{1}{6}$
• C. $\frac{1}{7}$
• D. $\frac{2}{7}$

Hint:

For “without replacement” probability, either multiply sequential probabilities or use the combinations formula $\frac{\binom{\text{favorable}}{r}}{\binom{\text{total}}{r}}$.

Answer 1

The Correct Option is B

- Step 1: Understanding the problem - We have 4 red balls and 5 blue balls in total. 
We want the probability that both balls drawn are red, without replacement. 
- Step 2: Total balls - $4 + 5 = 9$ balls in the bag. 
- Step 3: Probability first ball is red - \[ P(\text{first red}) = \frac{\text{Number of red balls}}{\text{Total balls}} = \frac{4}{9} \] 
- Step 4: After drawing one red ball - Now 3 red balls remain and total balls reduce to 8. 
- Step 5: Probability second ball is red - \[ P(\text{second red} \mid \text{first red}) = \frac{3}{8} \] 
- Step 6: Probability both red - Multiply the probabilities (since events are sequential and dependent): \[ P(\text{both red}) = \frac{4}{9} \times \frac{3}{8} = \frac{12}{72} = \frac{1}{6} \] 
- Step 7: Alternate method using combinations - Ways to choose 2 red balls = $\binom{4}{2} = 6$. 
Total ways to choose any 2 balls from 9 = $\binom{9}{2} = 36$. 
Probability = $\frac{6}{36} = \frac{1}{6}$. 
- Step 8: Conclusion - Probability = $\frac{1}{6}$, which is option (2).