Q1. [1]
From the data 1, 4, 7, 9, 16, 21, 25, if all the even numbers are removed, then the probability of getting at random a prime number from the remaining is :
- (a) $\frac{2}{5}$
- (b) $\frac{1}{5}$
- (c) $\frac{1}{7}$
- (d) $\frac{2}{7}$
Previously asked in CBSE board exam
2024 30/1/1 Q8
Generated by claude-sonnet-4-6 · 2026-06-15 10:31 · grounding rag
Model Answer
After removing even numbers (4, 16), the remaining data is: 1, 7, 9, 21, 25 → 5 numbers.
Prime numbers among these: 7 only → 1 prime.
$$P(\text{prime}) = \frac{1}{5}$$
Answer: (b) $\dfrac{1}{5}$
Explanation
- Even numbers in the set: 4 and 16. Remove them; remaining = {1, 7, 9, 21, 25} — 5 elements.
- Check primes: 1 is not prime; 7 is prime; 9 = 3×3 (not prime); 21 = 3×7 (not prime); 25 = 5×5 (not prime).
- Only 1 favourable outcome (7) out of 5 → P = 1/5.
- Common mistake: treating 1 as prime or 9/21/25 as prime — avoid this.
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