JEE · Maths · Sets & Statistics doubts
12 doubts found
In a survey, 30 like tea, 25 like coffee, 10 like both — how many like at least one, and how do I avoid double counting?
Use the inclusion–exclusion principle: n(tea ∪ coffee) = n(tea) + n(coffee) − n(both).
One employee earns far more than the rest — should I describe salaries by the mean or the median?
Use the median.
Two classes have the same average marks — how can statistics show they're actually very different?
Compare their spread using standard deviation.
What does “Inclusion–exclusion” actually mean in Sets & Statistics, and why does it matter for JEE?
for two sets, n(A∪B) = n(A) + n(B) − n(A∩B) — you subtract the overlap so it isn't double-counted.
I keep getting “Inclusion–exclusion” questions wrong in Sets & Statistics. What's the trap?
adding the two group sizes and forgetting to remove the overlap.
What does “Measures of spread” actually mean in Sets & Statistics, and why does it matter for JEE?
standard deviation measures how far the data typically lie from the mean — small means tightly clustered, large means spread out.
I keep getting “Measures of spread” questions wrong in Sets & Statistics. What's the trap?
thinking two data sets with the same mean must be alike.
What is the difference between mean and median?
The mean is the arithmetic average (sensitive to outliers); the median is the middle value (robust to outliers) — the median better represents skewed data.
What is the difference between union (A∪B) and intersection (A∩B)?
The union is everything in A OR B (or both); the intersection is only what's in A AND B — the shared overlap.
When do I use n(A∪B) = n(A) + n(B) − n(A∩B) in Sets & Statistics, and what does it mean?
inclusion–exclusion for two sets.
How is Sets & Statistics tested in JEE, and what should I focus on?
Prioritise the core ideas: Inclusion–exclusion, Measures of spread.
What are the most common mistakes students make in Sets & Statistics?
elements in both sets get counted twice, so you must subtract the intersection: n(A∪B) = n(A)+n(B)−n(A∩B).