Two reds without replacement — Probability Aptitude Question
Two reds without replacement: A bag has 4 red and 6 blue balls. Two balls are drawn one after another without… Answer and worked solution on CodeKairo.
- Topic: Probability
- Section: Quantitative Aptitude
- Difficulty: Medium
- Time target: 50 seconds
Question
A bag has 4 red and 6 blue balls. Two balls are drawn one after another without replacement. What is the probability that both are red?
Options
- 3/25
- 2/15
- 4/25
- 1/5
Show the answer and the worked solution
Correct answer: B. 2/15
Worked solution
- P(first red) = 4/10. P(second red | first red) = 3/9.
- P(both red) = 4/10 × 3/9 = 12/90 = 2/15.
- Or C(4,2)/C(10,2) = 6/45 = 2/15.
Answer: 2/15.
The idea
Without replacement, update both numerator and denominator after each draw — or count combinations directly. With replacement it would be (4/10)² = 4/25.
Shortcut. Cancel before multiplying: 4/10 × 3/9 = 2/5 × 1/3 = 2/15.
Common trap. 4/25 — using (4/10)², the with-replacement answer; after one red is drawn only 3 reds remain among 9 balls.