1. Define outcomes before counting them
A random experiment gives a sample space; events are subsets of it.
Define outcomes before counting them. Decide whether outcomes are equally likely before using favourable-over-total counting.
JEE JEE Main and JEE Advanced · Mathematics
Construct a valid sample space, model events, select addition, multiplication, conditional, total-probability or Bayes reasoning correctly, and distinguish independence from mutual exclusivity.
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In short
Probability starts with a sample space and clearly defined events. Addition handles unions, multiplication handles intersections, conditional probability updates the sample space after information is known, and Bayes' theorem reverses a conditional direction through a partition.
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Verified against the current JEE Main 2026 syllabus and JEE Advanced 2026 syllabus.
Sources: JEE Main 2026 syllabus and JEE Advanced 2026 syllabus, both linked in the sources section below.
A random experiment gives a sample space; events are subsets of it.
Define outcomes before counting them. Decide whether outcomes are equally likely before using favourable-over-total counting.
"Or" is union, "and" is intersection, "given" is conditioning.
Translate "or" to union, "and" to intersection and "given" to conditioning before selecting a formula.
Check whether events are mutually exclusive or independent rather than assuming either.
If A and B are mutually exclusive and both have positive probability, then P(A∩B)=0 while P(A)P(B)>0, so they are not independent. This distinction must remain explicit.
Conditional probability requires a positive denominator.
Check denominators before conditional probability, and for Bayes' theorem verify that the conditioning cases form an appropriate partition.
Every final probability must lie in [0,1].
Final probability must lie in [0,1]. A result outside this range signals a modelling or arithmetic error.
P of A complement equals one minus P of A.
Probability of the complement of A.
Use when — "Not A" questions.
Common trap — Complementing the wrong event.
P of A union B equals P of A plus P of B minus P of A intersection B.
Probability of the union of two events.
Use when — "A or B" questions.
Common trap — Adding the intersection twice.
P of A union B equals P of A plus P of B, valid only when A intersection B is empty.
Union probability for disjoint events.
Use when — Disjoint events.
Common trap — Using this for overlapping events.
P of A given B equals P of A intersection B divided by P of B, requiring P of B greater than zero.
Probability of A given B has occurred.
Use when — "A given B" questions.
Common trap — Zero denominator.
P of A intersection B equals P of B times P of A given B.
Intersection probability via conditioning on B.
Use when — Sequential or intersection events, with P(B)>0.
Common trap — Assuming independence.
P of A intersection B equals P of A times P of B given A.
Intersection probability via conditioning on A.
Use when — Same as multiplication rule, conditioning on A, with P(A)>0.
Common trap — Zero denominator in the conditional.
P of A intersection B equals P of A times P of B.
Definition of independence for events A and B.
Use when — Testing independence.
Common trap — Confusing independence with disjointness.
P of A equals the sum over i of P of A given B sub i times P of B sub i.
Total probability via a partition of the sample space.
Use when — Combining conditional routes.
Common trap — Cases not exhaustive or overlapping.
P of B sub j given A equals P of A given B sub j times P of B sub j, divided by the sum over i of P of A given B sub i times P of B sub i.
Reverses the conditioning direction through a partition.
Use when — Reverse conditioning.
Common trap — Missing prior probabilities.
Each p sub i is nonnegative and the sum of all p sub i equals one.
Validity condition for a discrete probability distribution.
Use when — Validate a random-variable table.
Common trap — Probabilities not summing to 1.
E of X equals the sum of x sub i times p sub i.
Expected value of a finite discrete random variable.
Use when — Advanced random-variable linkage.
Common trap — Averaging x-values without probabilities.
Variance of X equals the sum of x sub i minus mu, squared, times p sub i, which must be nonnegative.
Variance of a finite discrete random variable.
Use when — Advanced random-variable linkage.
Common trap — Negative result from arithmetic error.
Answer: 2/3
Known: P(A)=P(B)=1/2, P(R|A)=2/3, P(R|B)=1/3. A and B form a disjoint exhaustive partition of the box choice.
The question reverses the known conditional direction, so Bayes' theorem is appropriate.
P(A|R) = [(2/3)(1/2)] / [(2/3)(1/2)+(1/3)(1/2)] = (1/3)/(1/2) = 2/3.
Validity check: the answer lies in [0,1]. Observing red should favour Box A because Box A has the larger red proportion, so a posterior above 1/2 is directionally sensible.
Assuming outcomes are equally likely without justification.
Needs reviewWhy it happens
Favourable-over-total counting only applies when outcomes are actually equally likely.
How it is corrected
Justify the equally-likely assumption before using favourable-over-total counting.
Confusing union with intersection.
Knowledge gapWhy it happens
"Or" and "and" translate to different set operations with different formulas.
How it is corrected
Translate the wording into union or intersection language before selecting a formula.
Treating mutually exclusive events as independent.
Knowledge gapWhy it happens
If both events have positive probability, mutually exclusive events cannot be independent.
How it is corrected
Check both relationships separately rather than assuming one implies the other.
Cancelling conditional probabilities without a positive denominator.
Execution errorWhy it happens
Conditional probability is undefined when the conditioning event has probability zero.
How it is corrected
Check that the denominator is positive before using conditional probability.
Using Bayes without an exhaustive disjoint case structure.
Decision / selection errorWhy it happens
Bayes' theorem requires the conditioning cases to form a valid partition.
How it is corrected
Verify the partition is pairwise disjoint and exhaustive before applying Bayes' theorem.
Omitting prior probabilities.
Recall gapWhy it happens
Bayes' theorem requires the prior probability of each partition case.
How it is corrected
List all prior probabilities before computing the Bayes numerator and denominator.
Computing a random-variable mean before validating the distribution.
Decision / selection errorWhy it happens
An invalid distribution makes any computed mean meaningless.
How it is corrected
Check pᵢ≥0 and Σpᵢ=1 before computing expectation or variance.
Producing a probability outside [0,1].
Execution errorWhy it happens
This signals a modelling or arithmetic error somewhere in the solution.
How it is corrected
Validate the final answer lies in [0,1] as a routine check.
Official syllabus scope verified against NTA JEE Main 2026 and JEE Advanced 2026 documents. No weightage, frequency or PYQ-count claim is made.
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