Probability is one of those topics where students either love it or hate it. The ones who hate it usually tried to memorise formulas without understanding the underlying logic. Once the logic clicks, this becomes a reliable scoring topic across CBSE Class 10, Class 12, and JEE.
What Is Probability?
Probability measures how likely an event is to occur. Formally, it’s a number between 0 and 1.
- P = 0 means the event is impossible.
- P = 1 means the event is certain.
- Anything in between is a likelihood.
P(A) = Number of favourable outcomes / Total number of outcomes
This formula assumes all outcomes are equally likely. That’s an important assumption — a fair coin, an unbiased die, well-shuffled cards.
Key Vocabulary You Must Know
Sample Space (S): The set of all possible outcomes. Rolling a die: S = {1, 2, 3, 4, 5, 6}.
Event: A subset of the sample space. “Getting an even number” = {2, 4, 6}.
Complementary Event: If P(A) is the probability of event A, then P(A’) = 1 − P(A). This is the probability that A does NOT happen.
Mutually Exclusive Events: Two events are mutually exclusive if they can’t happen simultaneously. Getting a Head and getting a Tail on one coin flip are mutually exclusive.
Exhaustive Events: Events that cover the entire sample space. Getting H or T on a coin flip — one of these must happen.
Addition Rule of Probability
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
The subtraction of P(A ∩ B) prevents double-counting outcomes that belong to both A and B.
Special case: If A and B are mutually exclusive, P(A ∩ B) = 0, so: P(A ∪ B) = P(A) + P(B)
Example: A card is drawn from a standard deck of 52. What is the probability that it is a King or a Heart?
P(King) = 4/52. P(Heart) = 13/52. P(King AND Heart) = 1/52 (King of Hearts). P(King OR Heart) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13
Multiplication Rule of Probability
P(A ∩ B) = P(A) × P(B|A) = P(B) × P(A|B)
Special case: If A and B are independent (one event doesn’t affect the other), then: P(A ∩ B) = P(A) × P(B)
Example: A bag has 3 red and 2 blue balls. Two balls are drawn with replacement. Find the probability both are red.
P(first red) = 3/5. P(second red) = 3/5 (with replacement, so independent). P(both red) = 3/5 × 3/5 = 9/25
“With replacement” → independent events → just multiply probabilities directly. “Without replacement” → dependent events → use conditional probability.
Conditional Probability
This is where Class 12 probability begins. The conditional probability P(A|B) means: “Given that B has already occurred, what is the probability of A?”
P(A|B) = P(A ∩ B) / P(B), provided P(B) ≠ 0
Why this formula makes sense: When we know B has occurred, the sample space shrinks from S to B. We want to find what fraction of B also contains A. That’s exactly P(A ∩ B) / P(B).
Example: Two cards are drawn without replacement from a deck. Given that the first card is a King, what is the probability that the second is also a King?
After drawing one King, there are 3 Kings left out of 51 remaining cards. P(second King | first King) = 3/51 = 1/17
Independence vs. Mutual Exclusivity
Students confuse these two. They are almost opposite concepts.
| Property | Meaning | Formula |
|---|---|---|
| Independent | One doesn’t affect the other | P(A ∩ B) = P(A)·P(B) |
| Mutually exclusive | Both can’t happen together | P(A ∩ B) = 0 |
If two events have non-zero probability and are mutually exclusive, they cannot be independent. Because if A happens, B definitely doesn’t — so they affect each other.
Don’t say “A and B are independent because they are mutually exclusive.” That’s backwards. Mutually exclusive events with positive probabilities are actually dependent.
Total Probability Theorem
When you can’t directly compute P(A) but you can compute it across different cases (a “partition”), the Total Probability Theorem helps.
If B₁, B₂, …, Bₙ partition the sample space (mutually exclusive, exhaustive), then:
P(A) = P(A|B₁)P(B₁) + P(A|B₂)P(B₂) + … + P(A|Bₙ)P(Bₙ)
Example: Factory A produces 60% of items, Factory B produces 40%. 2% of A’s items are defective, 3% of B’s are defective. What is the probability that a randomly chosen item is defective?
P(Defective) = P(D|A)·P(A) + P(D|B)·P(B) = 0.02 × 0.60 + 0.03 × 0.40 = 0.012 + 0.012 = 0.024 = 2.4%
Bayes’ Theorem
Bayes’ theorem answers the reverse question: given that A has occurred, what is the probability it came from cause Bᵢ?
P(Bᵢ|A) = [P(A|Bᵢ) × P(Bᵢ)] / P(A)
where P(A) is found using the Total Probability Theorem
Terminology:
- P(Bᵢ) = Prior probability (before seeing the evidence)
- P(A|Bᵢ) = Likelihood (probability of evidence given the cause)
- P(Bᵢ|A) = Posterior probability (updated probability after evidence)
Example using the factory problem above: A defective item is found. What is the probability it came from Factory A?
P(A|Defective) = P(Defective|A) × P(A) / P(Defective) = (0.02 × 0.60) / 0.024 = 0.012 / 0.024 = 0.5 = 50%
Even though Factory A produces more items, both factories contribute equally to the defective pile.
Solved Examples
Example 1 (Easy): Two dice are rolled. Find P(sum = 7).
Total outcomes = 36. Favourable pairs: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — that’s 6 pairs. P(sum = 7) = 6/36 = 1/6
Example 2 (Medium): A bag has 5 red, 3 blue, 2 green balls. One ball is drawn. Find P(not red).
P(red) = 5/10 = 1/2. P(not red) = 1 − 1/2 = 1/2.
Or directly: 5 non-red balls out of 10. P = 5/10 = 1/2.
Example 3 (Hard): Three coins are tossed. Find P(at least one head).
Complement method: P(at least one H) = 1 − P(no heads) = 1 − P(all tails). P(all tails) = 1/2 × 1/2 × 1/2 = 1/8. P(at least one H) = 1 − 1/8 = 7/8
Example 4 (Medical Test — Bayes): A test for a disease is 95% accurate. The disease affects 1% of the population. If you test positive, what is the probability you have the disease?
Let D = has disease, D’ = no disease, T = test positive.
P(D) = 0.01, P(D’) = 0.99 P(T|D) = 0.95 (true positive rate) P(T|D’) = 0.05 (false positive rate)
P(T) = P(T|D)·P(D) + P(T|D’)·P(D’) = 0.95 × 0.01 + 0.05 × 0.99 = 0.0095 + 0.0495 = 0.059
P(D|T) = (0.95 × 0.01) / 0.059 = 0.0095 / 0.059 ≈ 0.161 = 16.1%
This counterintuitive result is Bayes’ theorem in action. A 95% accurate test still has only a ~16% chance of a true positive when the disease is rare. This is why medical screening requires multiple tests.
Exam-Specific Tips
CBSE Class 10:
- Classical probability from dice, cards, and bags are the standard question types.
- Always list or count the sample space for dice and coin problems.
- Complementary probability (1 − P) is a 2-mark question that shows up every year.
CBSE Class 12:
- Conditional probability, independence, Bayes theorem, and random variables form the bulk of the probability chapter.
- For Bayes theorem questions, set up a neat table with prior, likelihood, and joint probability columns. It reduces errors.
- Random variables and binomial distribution are related topics in the same unit.
JEE Main:
- Combinatorics and probability interlink heavily. Counting techniques (permutations, combinations) feed into probability calculations.
- Questions often involve “at least” or “at most” conditions — always think about the complement.
- Geometric probability occasionally appears — length or area as the sample space.
5 Common Mistakes
Mistake 1: Not listing outcomes carefully for two dice The sample space for two dice has 36 outcomes, not 11 (the possible sums from 2 to 12). Always count ordered pairs: (1,6) and (6,1) are different outcomes.
Mistake 2: Treating dependent events as independent Drawing without replacement changes the probabilities at each step. P(second red ball) ≠ P(first red ball) when there’s no replacement.
Mistake 3: Forgetting the complement is powerful “At least one” and “more than zero” questions are almost always easier with the complement: P = 1 − P(none at all).
Mistake 4: Writing P(A) > 1 Probability is always between 0 and 1, inclusive. If your calculation gives 1.2 or −0.3, something is wrong. Check your sample space and favourable outcome counts.
Mistake 5: Misidentifying P(A|B) and P(B|A) These are completely different. P(Rain|Dark clouds) ≠ P(Dark clouds|Rain). In Bayes’ theorem, clearly label what is given and what you’re finding before substituting.
Real-World Examples
Example 1: COVID-19 Rapid Antigen Tests and False Positives
During the 2021 COVID wave, ICMR-approved rapid antigen tests had roughly 70% sensitivity and 99% specificity. Suppose 1% of Mumbai’s population was actually infected on a given day. If your test comes back positive, what’s the real chance you have COVID? Most students instinctively say “99%.” Bayes’ Theorem tells a different story: . Nearly six out of ten positive results were false alarms.
Connect to the syllabus: This is a direct application of Bayes’ Theorem, , which appears in CBSE Class 12 Chapter 13 and JEE Main under conditional probability.
Example 2: The IPL Toss and Match-Winning Probability
In IPL 2023, teams winning the toss at Chepauk (Chennai) chose to field first in 18 of 20 matches — and won 14 of those 18. If we treat each match independently, the probability that a toss-winning, field-first team wins on this ground is . A team captain using this historical frequency as a prior is essentially doing frequentist probability estimation. Fantasy-league players on Dream11 quietly use these numbers every match day.
Connect to the syllabus: This illustrates empirical (statistical) probability, , the foundational definition in CBSE Class 10 and the starting point for JEE’s probability chapter.
Example 3: Quality Control at a Tata Steel Plant
A Tata Steel rolling mill produces steel rods where, historically, 2% are defective. Two independent inspectors each catch a defective rod with 90% accuracy. What’s the probability that a defective rod escapes both inspectors? Since the checks are independent, . So only of all rods — 2 per 10,000 — slip through undetected. This is why Six Sigma manufacturing uses layered independent checks.
Connect to the syllabus: This uses the multiplication rule for independent events, , a core result tested in both CBSE Class 12 boards and JEE Main numerical problems.
Practice Questions
Q1. A die is rolled. Find the probability of getting a number greater than 4.
Favourable outcomes: {5, 6} — that’s 2 outcomes. Total = 6.
P = 2/6 = 1/3
Q2. Two dice are rolled. Find P(sum is even).
Sum is even when both dice are even or both are odd. P(both even) = (3/6)(3/6) = 1/4. P(both odd) = (3/6)(3/6) = 1/4. P(even sum) = 1/4 + 1/4 = 1/2
Q3. A card is drawn from a deck. Find P(Jack or Black card).
P(Jack) = 4/52. P(Black) = 26/52. P(Jack AND Black) = 2/52 (black Jacks). P(Jack OR Black) = 4/52 + 26/52 − 2/52 = 28/52 = 7/13
Q4. If P(A) = 1/3, P(B) = 1/4, and A, B are independent. Find P(A ∪ B).
P(A ∩ B) = P(A)·P(B) = 1/3 × 1/4 = 1/12 P(A ∪ B) = 1/3 + 1/4 − 1/12 = 4/12 + 3/12 − 1/12 = 6/12 = 1/2
Q5. A bag has 4 white and 6 black balls. Two balls are drawn without replacement. Find P(both white).
P(first white) = 4/10. P(second white | first white) = 3/9. P(both white) = 4/10 × 3/9 = 12/90 = 2/15
Q6. Three coins are tossed. Find P(exactly 2 heads).
Sample space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} = 8 outcomes.
Exactly 2 heads: {HHT, HTH, THH} = 3 outcomes.
P = 3/8
Q7. P(A) = 0.6, P(B) = 0.5, P(A ∩ B) = 0.3. Are A and B independent?
For independence: P(A ∩ B) must equal P(A)·P(B) = 0.6 × 0.5 = 0.3. Since P(A ∩ B) = 0.3 = 0.3, yes, A and B are independent.
Q8. Bag 1 has 3 red, 2 blue. Bag 2 has 2 red, 4 blue. One bag is chosen randomly, then one ball is drawn. Find P(red ball).
P(Bag 1) = P(Bag 2) = 1/2. P(Red|Bag 1) = 3/5. P(Red|Bag 2) = 2/6 = 1/3. P(Red) = (3/5)(1/2) + (1/3)(1/2) = 3/10 + 1/6 = 9/30 + 5/30 = 14/30 = 7/15
Frequently Asked Questions
What is the difference between theoretical and experimental probability? Theoretical probability is calculated mathematically (assuming ideal conditions). Experimental probability is based on actual trials. As the number of trials increases, experimental probability approaches theoretical probability — this is the Law of Large Numbers.
When should I use combinations vs. direct counting for probability? When the problem involves selecting objects from a group and order doesn’t matter, combinations (ⁿCᵣ) are cleaner. For dice, coins, and sequences where order matters, direct counting is usually simpler.
Is Bayes theorem in CBSE Class 10? No. Class 10 covers only basic/classical probability. Conditional probability and Bayes theorem are Class 12 topics. Don’t mix them up.
How do I verify my answer in probability? The sum of probabilities of all outcomes must equal 1. For complementary events, P(A) + P(A’) = 1. Use these as a sanity check.
What is a probability distribution? A probability distribution lists all possible values of a random variable and their associated probabilities. This is a Class 12 / JEE topic — binomial distribution being the most commonly tested.
Can two events both be independent AND mutually exclusive? Only if at least one of them has probability 0. If P(A) > 0 and P(B) > 0, then mutually exclusive events cannot be independent (because knowing one occurred tells you the other didn’t).