Probability

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Probability — Quick Facts

Basic Probability: $$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$$

Key Properties:

• $0 \leq P(E) \leq 1$
• $P(S) = 1$ (sample space)
• $P(\emptyset) = 0$ (empty set)
• $P(E’) = 1 - P(E)$ (complement)

Addition Rule: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

For mutually exclusive events: $P(A \cup B) = P(A) + P(B)$

Multiplication Rule: $$P(A \cap B) = P(A) \times P(B|A)$$

For independent events: $P(A \cap B) = P(A) \times P(B)$

⚡ CAT Exam Tip: Use the complement method for “at least one” problems: $P(\text{at least one}) = 1 - P(\text{none})$.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding.

Probability — Study Guide

Combinations in Probability:

$$C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}$$

Example: From a pack of 52 cards, 3 cards are drawn. Find probability of getting exactly 2 hearts.

Number of ways to choose 2 hearts from 13: $C(13, 2) = 78$ Number of ways to choose 1 non-heart from 39: $C(39, 1) = 39$ Favourable outcomes: $78 \times 39 = 3042$

Total ways to choose 3 cards from 52: $C(52, 3) = 22100$

$P(\text{2 hearts}) = \frac{3042}{22100} = \frac{1521}{11050} \approx 0.138$

Conditional Probability:

$P(A|B)$ — probability of A given B has occurred: $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$

Example: In a class, 60% students play cricket, 45% play football, and 30% play both. If a student plays cricket, what’s the probability they also play football?

$P(\text{cricket and football}) = 0.30$ $P(\text{cricket}) = 0.60$

$P(\text{football} | \text{cricket}) = \frac{0.30}{0.60} = 0.5$

Bayes’ Theorem: $$P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}$$

⚡ Common Student Mistake: Confusing $P(A|B)$ with $P(B|A)$. Read the question carefully — “given that A happened, what’s the probability of B?”

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Probability — Comprehensive Notes

Permutation-Based Probability:

$$P(n, r) = \frac{n!}{(n-r)!}$$

Example: How many 4-digit numbers can be formed from {1, 2, 3, 4, 5} with no repetition? What’s the probability such a number is even?

Total 4-digit numbers: $5 \times 4 \times 3 \times 2 = 120$

For even numbers, last digit must be {2, 4} = 2 choices First digit: 4 choices (cannot be 0) Second digit: 3 choices Third digit: 2 choices Favourable: $4 \times 3 \times 2 \times 2 = 48$

$P(\text{even}) = \frac{48}{120} = 0.4$

Derangements (No Fixed Points):

The number of ways $n$ items can be arranged with no item in its original position: $$D_n = n! \left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + … + \frac{(-1)^n}{n!}\right)$$

Example: 3 letters in 3 envelopes, none correctly placed. In how many ways?

$D_3 = 3!(1 - 1 + \frac{1}{2} - \frac{1}{6}) = 6 \times \frac{1}{3} = 2$

Expected Value:

$$E(X) = \sum x_i \cdot P(x_i)$$

Example: A game costs ₹100 to play. You win ₹300 with probability 0.2, ₹100 with probability 0.5, and lose otherwise. Find expected profit.

$E(\text{profit}) = 200(0.2) + 0(0.5) + (-100)(0.3) = 40 - 30 = ₹10$

Since $E > 0$, the game is favourable.

Binomial Probability:

$$P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}$$

Example: A coin is tossed 5 times. Find probability of getting exactly 3 heads.

$n = 5$, $r = 3$, $p = 0.5$ $P(X = 3) = \binom{5}{3} (0.5)^3 (0.5)^2 = 10 \times 0.125 \times 0.25 = 0.3125$

JAMB Pattern Analysis (CAT 2015-2024):

• 2015: Basic probability with cards
• 2017: Conditional probability with Venn diagrams
• 2019: Binomial distribution
• 2021: Expected value in word problems
• 2023: Derangements
• 2024: Bayes’ theorem application

⚡ Exam Strategy: For complex probability problems, use a tree diagram. Branch for each event, label probabilities, multiply along branches for “AND,” add for alternative paths.