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Probability: Advanced revision notes

Every idea in the Probability: Advanced chapter for JEE Main, with the usual mistakes and the fast methods. Topic 1 is open to everyone; sign in free to read all 16 and download the PDF.

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How we count: 7× asked means 7 past JEE Main questions in the PrepWiser bank needed that concept, and "last" is the latest year one appeared. ★ marks ideas asked 8 or more times.

Conditional probability and multiplication theorem

Conditional probability
7× asked, last 2025
P(A∣B)=P(A∩B)P(B),P(A∩B)=P(B) P(A∣B)=P(A) P(B∣A)P(A\mid B)=\frac{P(A\cap B)}{P(B)},\qquad P(A\cap B)=P(B)\,P(A\mid B)=P(A)\,P(B\mid A)
Needs P(B)>0P(B)>0. BB becomes the new sample space; for equally likely outcomes it is n(A∩B)/n(B)n(A\cap B)/n(B).
4484888812112233445566first dienew sample space:9 cells (sum 4, 8, 12)a 4 shows: only (4,4)(4,4)P=19P=\frac19
Given "sum is a multiple of 4", keep only those 9 outcomes, then count the ones with a 4
  • From two conditionals and P(A∩B)P(A\cap B): P(A)=P(A∩B)/P(B∣A)P(A)=P(A\cap B)/P(B\mid A) and P(B)=P(A∩B)/P(A∣B)P(B)=P(A\cap B)/P(A\mid B). Then P(A∪B)P(A\cup B) by addition, and P(A′∩B′)=1−P(A∪B)P(A'\cap B')=1-P(A\cup B).
  • Not P(B)P(B) on top for P(B∣A∪B′)P(B\mid A\cup B'). The top is P(B∩(A∪B′))=P(A∩B)P(B\cap(A\cup B'))=P(A\cap B), because B∩B′B\cap B' is empty. The bottom is P(A)+P(B′)−P(A∩B′)P(A)+P(B')-P(A\cap B').
  • Quick method. Reduced sample space: list only the outcomes of BB, then count those in AA. Two dice, sum a multiple of 4: 9 outcomes, and only (4,4)(4,4) shows a 4, so the chance of a 4 is 1/91/9.
Complements inside a condition
3× asked, last 2022
P(A′∣B)=1−P(A∣B),P(A∣B′)=P(A)−P(A∩B)1−P(B)P(A'\mid B)=1-P(A\mid B),\qquad P(A\mid B')=\frac{P(A)-P(A\cap B)}{1-P(B)}
Complement the event, never the condition: P(A∣B′)P(A\mid B') is not 1−P(A∣B)1-P(A\mid B).
AABBSSA∩B′A\cap B'B′B' (all shaded)
P(A∣B′)P(A\mid B'): red part over all shaded; the top is P(A)−P(A∩B)P(A)-P(A\cap B), not P(A)P(A)
  • Not P(A)/P(B′)P(A)/P(B') for P(A∣B′)P(A\mid B'). The top must be P(A∩B′)=P(A)−P(A∩B)P(A\cap B')=P(A)-P(A\cap B). Find P(A∩B)P(A\cap B) first, from P(A∪B)P(A\cup B) by the addition rule.
Multiplication theorem
2× asked, last 2025
P(A∩B∩C)=P(A) P(B∣A) P(C∣A∩B)P(A\cap B\cap C)=P(A)\,P(B\mid A)\,P(C\mid A\cap B)
Take the events in time order. Each factor is the chance at that step, given all the earlier steps.
23\frac{2}{3}not intercepted34\frac{3}{4}hit: 12\frac{1}{2}14\frac{1}{4}miss: 16\frac{1}{6}13\frac{1}{3}intercepted: 13\frac{1}{3}
A hit needs both steps: P(N) P(H∣N)=23⋅34=12P(N)\,P(H\mid N)=\frac23\cdot\frac34=\frac12; the 34\frac34 alone is the trap
  • Not the given conditional as the answer. 'hits, given not intercepted' is P(H∣N)P(H\mid N). A hit needs both steps: P(N) P(H∣N)P(N)\,P(H\mid N), P(N)=2/3P(N)=2/3, P(H∣N)=3/4P(H\mid N)=3/4 give 1/21/2.
  • Quick method. Without replacement, the kk-th draw has the same chance as the first. From 4 white, 6 black: both black is (6/10)(5/9)=1/3(6/10)(5/9)=1/3, so P(1st B∣2nd B)=(1/3)/(6/10)=5/9P(\text{1st B}\mid\text{2nd B})=(1/3)/(6/10)=5/9.

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Practise Probability: Advanced

13 more concepts in 3 topics

Independent events and repeated trials, Total probability and Bayes' theorem and Random variables and distributions. Read the first part free. Sign in free with Google to read the rest and download the PDF.

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Independent events and repeated trials

  • Independent events4×
  • Three events: pairwise vs mutual3×
  • Waiting for the first success3×

Total probability and Bayes' theorem

  • Total probability theorem2×
  • Transfer, then draw2×
  • Bayes' theorem14×
  • Unknown make-up and lost cards4×

Random variables and distributions

  • Probability distribution10×
  • Mean and variance7×
  • Linear change of a variable1×
  • Distributions from experiments5×
  • Sample without replacement7×
  • Binomial distribution (Bernoulli trials)23×

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