Q.7 From the set of 10 numbers {1, 2,…,10} three numbers are selected at random without replacement. The probability that the sum of these selected numbers is 9, is
The probability that three numbers selected at random without replacement from {1, 2, …, 10} sum to 9 is 3/120 = 1/40.
Total Outcomes
The total number of ways to choose 3 distinct numbers from 10 is the combination C(10,3) = 10!/(3!(10-3)!) = (10×9×8)/(3×2×1) = 120.
Favorable Outcomes
The only combinations summing to 9 are (1,2,6), (1,3,5), and (2,3,4).
No other sets of three distinct numbers from 1 to 10 sum exactly to 9, as larger numbers exceed the target when paired with smaller ones.
Probability Calculation
Probability equals favorable outcomes divided by total outcomes: 3/120 = 1/40.
Option Analysis
| Option | Value | Reason |
|---|---|---|
| (A) 1/40 | Correct | Matches 3/120 |
| (B) 1/20 | Incorrect | Equals 6/120, overcounts favorable cases |
| (C) 3/10 | Incorrect | Far too high (36/120), ignores combinations |
| (D) 3/80 | Incorrect | Assumes smaller sample space like C(8,3)=56, but total is 120 |
CSIR NET Exam Context
The probability three numbers from 1 to 10 sum to 9 is a classic combinations problem for CSIR NET Life Sciences and math competitive exams. Selecting three distinct numbers without replacement from the set {1, 2, …, 10} requires computing favorable outcomes over total possibilities.
Step-by-Step Solution
- Total ways: C(10,3) = 120
- Favorable: Only (1,2,6), (1,3,5), (2,3,4) sum to 9
- Thus, P = 3/120 = 1/40, option (A)
This reinforces combination formulas essential for competitive exams like CSIR NET.
