Probability
Without Replacement — P(a−b≥10), m+n
nta_pyq_2026_jan
Grade 12
Question:
From the first 100 natural numbers, two numbers first $a$ and then $b$ are selected randomly without replacement. If the probability that $a-b\geq10$ is $\dfrac{m}{n}$, $\gcd(m,n)=1$, then $m+n$ is equal to _____.
Step-by-Step Solution
Key Concept: Total ordered selections $=100\times99=9900$. Favorable: for each $b$ from 1 to 90, $a$ ranges from $b+10$ to 100, giving $91-b$ choices.
$P=\tfrac{91}{220}$. $m+n=311$.
Correct Answer: 311