Probability
Probability
nta_abhyas_2025
Grade 12
Question:
If $4$ distinct numbers are chosen randomly from the first $100$ natural numbers, then the probability that all $4$ of them are either divisible by $3$ or divisible by $5$ is
\frac{2625}{3921225}
\frac{2925}{3921225}
\frac{2825}{3921225}
\frac{2725}{3921225}
Step-by-Step Solution
Key Concept: Use the inclusion-exclusion principle to count elements satisfying at least one condition from multiple sets.
Total numbers from 1 to 100 is $^{100}C_1 = 100$. Numbers divisible by 3: $n(3) = 33$. Numbers divisible by 5: $n(5) = 20$. Numbers divisible by both 3 and 5: $n(3 \cap 5) = 6$. By inclusion-exclusion: $33 + 20 - 6 = 47$. Required probability = $\frac{^{47}C_1}{^{100}C_1} = \frac{47}{100}$.
Correct Answer: 47