UPSC Prelims 2007 Q139: Science & Technology – Quantitative Aptitude
Three dice (each having six faces with each face showing one number from 1 to 6) are
rolled. What is the number of possible outcomes such that at least one die shows the number 2?
- (a) 36
- (b) 81
- (c) 91
- (d) 116
Answer: (c) 91
Explanation
Three dice — at least one shows 2
Use the complement method.
| Step | Calculation | Value |
|---|---|---|
| All outcomes | 6 × 6 × 6 | 216 |
| Outcomes with no 2 | 5 × 5 × 5 | 125 |
| At least one 2 | 216 − 125 | 91 |
Check by cases:
| Exactly | Ways |
|---|---|
| One 2 | 3 × 5 × 5 = 75 |
| Two 2s | 3 × 5 = 15 |
| Three 2s | 1 |
| Total | 91 |
Answer: (c) 91.
Why each option is right or wrong
- (a) Incorrect. 36 is the number of outcomes for two dice.
- (b) Incorrect. 81 has no valid basis here.
- (c) CORRECT. 216 − 125 = 91 outcomes with at least one six.
- (d) Incorrect. 116 over-counts; the complement method gives 91.
Key point
'At least one' = Total − None. 6³ − 5³ = 216 − 125 = 91.
Remember
At least one 2 on three dice = 216 − 125 = 91.
Common mistakes
- Doing 3 × 36 = 108 — this counts outcomes with two or three 2s more than once.
- Forgetting the complement shortcut.
- Treating the dice as identical — they are distinct, so 216 total outcomes.
Exam tip
Whenever you see 'at least one', think Total − None.
Sources
- NCERT Class XI — Mathematics, Ch 16 (Probability)
- R. S. Aggarwal — Quantitative Aptitude for Competitive Examinations (Probability)