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Probability — JEE Mains 2026 Revision capsule 

A compact, high-yield probability capsule — every formula, every trap, P&C connections, solved examples and a fast recall ladder. MathJax renders all equations.

🔁 Revision Series 🔗 Permutation & Combination 🔗 Binomial

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Axioms
\(0\le P(A)\le1,\; P(S)=1,\; P(A^c)=1-P(A)\)

Add/Mul Rules
\(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
\(P(A\cap B)=P(A)P(B|A)\)

Bayes & Total
\(P(A)=\sum_i P(B_i)P(A|B_i)\)
\(P(B_k|A)=\dfrac{P(B_k)P(A|B_k)}{\sum_i P(B_i)P(A|B_i)}\)

Exam-Killer (memorize):

• “At least one” → use complement: \(1-P(\text{none})\).
• Without replacement → events dependent. Use combinations, not product of probabilities.
• Bayes = prior × likelihood / evidence — draw tree for clarity.

1. Core Definitions

Experiment, Sample Space \(S\), Event \(E\). Fundamental formula:

Classical probability:
\[ P(E)=\frac{n(E)}{n(S)} \]

Types of events

• Sure event: \(P=1\). Impossible: \(P=0\).
• Mutually exclusive: \(A\cap B=\varnothing\).
• Complement: \(P(A^c)=1-P(A)\).

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2. Addition & Multiplication Rules

Addition (general):
\[ P(A\cup B)=P(A)+P(B)-P(A\cap B) \] For disjoint events → \(P(A)+P(B)\).

Multiplication (chain):
\[ P(A\cap B)=P(A)\,P(B|A)=P(B)\,P(A|B) \] If independent → \(P(A\cap B)=P(A)P(B)\).

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3. Conditional Probability & Bayes

Conditional:
\[ P(A|B)=\frac{P(A\cap B)}{P(B)}\quad(P(B)>0) \]

Total probability:
If \(B_1,\dots,B_n\) partition \(S\), \[ P(A)=\sum_{i=1}^n P(B_i)P(A|B_i) \]

Bayes' theorem:
\[ P(B_k|A)=\frac{P(B_k)P(A|B_k)}{\sum_{i=1}^n P(B_i)P(A|B_i)} \]

P&C Connection (how to count):

Use Permutation & Combination to compute both numerator (favourable) and denominator (total) — then apply \(P=\dfrac{n(E)}{n(S)}\).

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4. Bernoulli Trials & Binomial

Exactly \(k\) successes in \(n\) trials:
\[ P(X=k)=\binom{n}{k}p^k(1-p)^{\,n-k} \]

Useful variations:

• \(\)At least one success: \(1-(1-p)^n\).
• \(\)Odd/even heads: use binomial sum or symmetry shortcuts.

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5. Expectation & Useful Distributions

Expectation (discrete):
\[ E(X)=\sum_x x\,P(X=x) \]

Examples: coin(n tosses) → \(E(\text{heads})=np\). Fair die → \(E(X)=3.5\).

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6. JEE Traps & How to Avoid (Red Alerts)

• Dependent vs independent: Drawing without replacement is dependent — recompute totals after each draw.
• Equally likely assumption: verify before using \(n(E)/n(S)\).
• Conditional-set misread: Always restrict sample space when a condition is given — re-normalize probabilities.
• Overcounting/undercounting: check order vs selection, use combinations for unordered selection.

7. Fast Revision Ladder (10s recall)

Core
\(P=\dfrac{n(E)}{n(S)}\)

Cond.
\(P(A|B)=\dfrac{P(A\cap B)}{P(B)}\)

Binomial
\(\binom{n}{k}p^k(1-p)^{n-k}\)

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8. Solved Examples (Short & JEE-style)

Ex 1 — Two dice: probability sum divisible by 3

Solution (counting):

Sample space size \(=36\). Sums divisible by 3 are 3,6,9,12.

Count outcomes: sum=3→2, sum=6→5, sum=9→4, sum=12→1. Total favourable \(=2+5+4+1=12\).

Therefore \(P=\dfrac{12}{36}=\dfrac{1}{3}.\)

Ex 2 — Biased coin p=0.6, 5 tosses: P(exactly 3 heads)

Solution:

\[ P=\binom{5}{3}(0.6)^3(0.4)^2 = 10\times 0.216\times 0.16 = 0.3456 \]

So probability ≈ 0.3456.

Ex 3 — Two balls from (3R,4B,5G): P(both same color)

Solution:

Total ways to pick 2 balls: \(\binom{12}{2}=66\).

Favourable: RR: \(\binom{3}{2}=3\), BB: \(\binom{4}{2}=6\), GG: \(\binom{5}{2}=10\). Total \(=19\).

\(P=\dfrac{19}{66}\).

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9. Practice Problems (Try First)

1. Form a 3-digit number from digits 1–6 without repetition: P(it is even but not divisible by 4).
2. Two cards drawn without replacement from 52: P(both are hearts).
3. Box A has 2 defective in 10, Box B has 3 defective in 20. Choose a box at random and pick one item. P(item defective?) (use Bayes)
4. From 6 men and 4 women, committee of 4 chosen. P(at least 2 women).

Show Quick Answers / Hints

1. Count total \(6\cdot5\cdot4=120\) numbers; check last digit even (2,4,6) and test divisibility by 4 using last two digits.
2. \(\dfrac{13}{52}\times\dfrac{12}{51}=\dfrac{1}{17}\) (or use combinations: \(\dfrac{\binom{13}{2}}{\binom{52}{2}}\)).
3. Total P(def)= \( \tfrac12\cdot\tfrac{2}{10} + \tfrac12\cdot\tfrac{3}{20}=0.1 + 0.075=0.175\). For Bayes compute conditional probabilities accordingly.
4. Compute complement P(0 or 1 woman) and subtract from 1.

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10. One-line Takeaways

• Always restructure sample space after a condition; renormalize probabilities.
• Count using P&C first — probability is a ratio.
• For “at least one”, use complement — saves time and reduces errors.
• Bayes = reverse conditional probability; draw trees for clarity.

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