UPSC Prelims 2004 Q147: Science & Technology – Quantitative Aptitude
Three students are picked at random from a school having a total of 1000 students. The
probability that these three students will have identical date and month of their birth is
- (a) 3/1000
- (b) 3/365
- (c) 1/(365)²
- (d) 1/(365)³
Answer: (c) 1/(365)²
Explanation
Probability of identical birthdays among three students
| Step | Probability |
|---|---|
| First student | Any date: 1 |
| Second student has the same date and month | 1/365 |
| Third student has the same date and month | 1/365 |
| All three | 1 x 1/365 x 1/365 = 1/(365)^2 |
The first student can have any birthday; the other two must each match it with probability 1/365, so the probability is 1/(365)^2.
Answer: (c) 1/(365)^2.
Why each option is right or wrong
- (a) Incorrect. 3/1000 has no basis in the problem.
- (b) Incorrect. 3/365 is far too large.
- (c) CORRECT. (1/365) x (1/365) = 1/(365)^2.
- (d) Incorrect. 1/(365)^3 would fix the first student's date as well.
Key point
Birthday matches: fix one person, then multiply chances for each other.
Remember
The chance of three identical birthdays is 1/(365)^2.
Common mistakes
- Using 1/(365)^3 by fixing the first date too.
- Adding probabilities instead of multiplying.
- Using the school size of 1000 students.
Exam tip
Three people: (1/365) per extra person, so squared.
Sources
- NCERT Class XI Mathematics — Probability
- R. S. Aggarwal — Quantitative Aptitude for Competitive Examinations (Probability)