Why this chapter is counting with a division
Ask for the probability that a hand of five cards contains exactly two aces, and the work is entirely combinatorial: count the favourable hands, count the total hands, divide. The probability layer contributes one operation; Permutations and Combinations contributes everything else.
That is why students who found the previous chapter hard find this one harder. The remedy is not to study probability more, but to make the counting automatic first - and then to learn the handful of genuinely probabilistic ideas, which are conditional probability, independence and the standard distributions.
How JEE actually asks Probability
NTA publishes no chapter-wise weightage, so figures circulating online are coaching estimates from past papers; confirm your session's syllabus and pattern in the official NTA information bulletin.
Three families recur. Basic questions on dice, coins and cards test the addition and multiplication rules. Conditional probability and Bayes theorem appear regularly and are the most distinctive part of the chapter. The binomial distribution supplies short numerical questions, usually on exactly r successes or at least one success.
Key concepts, compressed
- ✓Classical probability is favourable outcomes over total outcomes, valid only when outcomes are equally likely.
- ✓The addition rule handles 'or': P(A ∪ B) = P(A) + P(B) − P(A ∩ B), with the subtraction removing the double count.
- ✓The multiplication rule handles 'and': P(A ∩ B) = P(A)P(B|A), which reduces to P(A)P(B) when the events are independent.
- ✓Conditional probability P(A|B) restricts the sample space to B, so the denominator becomes P(B) rather than 1.
- ✓Bayes theorem inverts a conditional, converting P(evidence given cause) into P(cause given evidence).
- ✓The binomial distribution counts successes in n independent trials with constant success probability p.
Formulas you need before attempting the questions
| Rule | Formula | Note |
|---|---|---|
| Classical probability | P(A) = favourable / total | equally likely outcomes |
| Complement | P(A′) = 1 − P(A) | |
| Addition rule | P(A ∪ B) = P(A) + P(B) − P(A ∩ B) | |
| Mutually exclusive | P(A ∩ B) = 0 | cannot both occur |
| Independent | P(A ∩ B) = P(A)P(B) | one does not affect the other |
| Conditional probability | P(A|B) = P(A ∩ B)/P(B) | |
| Multiplication rule | P(A ∩ B) = P(A)P(B|A) | |
| Bayes theorem | P(Aᵢ|B) = P(B|Aᵢ)P(Aᵢ) / Σ P(B|Aⱼ)P(Aⱼ) | |
| Binomial probability | P(X = r) = ⁿCr pʳ q^(n−r) | q = 1 − p |
| Binomial mean | np | |
| Binomial variance | npq | always less than the mean |
| At least one | 1 − P(none) |
The five mistakes that cost the most marks
- ✓Treating mutually exclusive and independent as the same thing. They are near-opposites: mutually exclusive events are maximally dependent, since one occurring guarantees the other did not.
- ✓Reversing the conditional. P(disease given a positive test) is not P(positive test given disease), and confusing them is the classic error Bayes theorem exists to fix.
- ✓Forgetting to subtract the intersection in the addition rule, which double counts the overlap.
- ✓Applying the binomial distribution to sampling without replacement. The trials are then neither independent nor constant in probability.
- ✓Assuming two outcomes means each has probability one half. Symmetry has to be justified, not assumed.
Practice set 1: basic probability
1. A fair die is rolled. What is the probability of an even number?
1/2. The favourable outcomes are 2, 4 and 6 - three of six equally likely results - so the probability is 3/6 = 1/2. The phrase 'fair die' is what licenses treating the outcomes as equally likely, and classical probability requires that assumption.
2. Two fair dice are rolled. What is the probability that the sum is 7?
1/6. There are 36 equally likely ordered outcomes. The sum of 7 arises from (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1) - six of them - so the probability is 6/36 = 1/6. Note that (1,6) and (6,1) are distinct outcomes; treating them as one is the standard error and would give 3/21.
3. For the same two dice, what is the probability that the sum is 8?
5/36. The favourable pairs are (2,6), (3,5), (4,4), (5,3) and (6,2) - five outcomes. Note (4,4) occurs only once, since both dice showing 4 is a single outcome. Seven is the most likely sum precisely because it has the most such pairs, which is why this pairing with question 2 is instructive.
4. A fair coin is tossed twice. What is the probability of two heads?
1/4. The sample space is HH, HT, TH, TT - four equally likely outcomes, one favourable. Equivalently, the tosses are independent so the probability is (1/2)(1/2) = 1/4. Both routes agree, and the second generalises to more tosses without enumeration.
5. A card is drawn from a standard deck. What is the probability it is a king?
1/13. There are 4 kings among 52 cards, giving 4/52 = 1/13. Standard deck facts worth having ready: 52 cards, 4 suits of 13, 26 red and 26 black, 12 face cards, and 4 of each rank.
6. If P(A) = 0.3, what is P(A′)?
0.7. The complement rule gives P(A′) = 1 − P(A) = 0.7. This is the most useful single relation in the chapter, because it converts every 'at least one' question into a much simpler 'none' question.
7. How many outcomes are in the sample space when two dice are rolled?
36. Each die has 6 outcomes and they are independent, so 6 × 6 = 36 ordered pairs. Getting the sample space right is the first step of any classical probability question, and treating the dice as indistinguishable - which would give 21 - is what causes the errors in questions 2 and 3.
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Practice set 2: addition and multiplication rules
8. Given P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.2, find P(A ∪ B).
0.7. The addition rule gives 0.5 + 0.4 − 0.2 = 0.7. Without subtracting the intersection you would get 0.9, which double counts the region where both occur. Every 'A or B' question needs this subtraction unless the events are mutually exclusive.
9. What does it mean for two events to be mutually exclusive?
That they cannot both occur, so P(A ∩ B) = 0 and the addition rule simplifies to P(A) + P(B). Rolling a 2 and rolling a 5 on a single die are mutually exclusive. The term describes the events themselves, not any numerical coincidence.
10. What does it mean for two events to be independent?
That the occurrence of one does not change the probability of the other, so P(A ∩ B) = P(A)P(B) and P(A|B) = P(A). Successive coin tosses are independent. Note this is a statement about influence, whereas mutual exclusivity is a statement about compatibility - which is why the next question matters.
11. Can two events be both mutually exclusive and independent?
Not unless at least one has probability zero. Mutual exclusivity forces P(A ∩ B) = 0, while independence requires P(A ∩ B) = P(A)P(B), and those agree only if one probability is zero. In fact mutually exclusive events are strongly dependent: knowing A occurred tells you with certainty that B did not.
12. A card is drawn. What is the probability it is a king or a heart?
4/13. There are 4 kings and 13 hearts, but the king of hearts is in both sets, so the count is 4 + 13 − 1 = 16, giving 16/52 = 4/13. Adding 4/52 and 13/52 without subtracting would give 17/52 and count that one card twice.
13. Two independent events have probabilities 0.6 and 0.5. What is the probability both occur?
0.3. For independent events the probabilities multiply: 0.6 × 0.5 = 0.3. If they were not independent you would need P(A)P(B|A) instead, and the answer could be anything from 0.1 to 0.5 depending on the dependence.
Practice set 3: conditional probability and Bayes theorem
14. For P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.2, find P(A|B).
0.5. Conditional probability is P(A|B) = P(A ∩ B)/P(B) = 0.2/0.4 = 0.5. The denominator is P(B) rather than 1 because conditioning restricts attention to the outcomes where B occurred - the sample space has shrunk.
15. Are A and B independent in that example?
Yes. P(A|B) = 0.5 which equals P(A) = 0.5, so knowing B occurred does not change the probability of A. Equivalently, P(A)P(B) = 0.5 × 0.4 = 0.2, which matches P(A ∩ B). Either test suffices, and both must give the same verdict.
16. State Bayes theorem and say what it is for.
P(Aᵢ|B) = P(B|Aᵢ)P(Aᵢ) divided by the sum of P(B|Aⱼ)P(Aⱼ) over all causes. It converts P(evidence given cause), which you usually know, into P(cause given evidence), which you usually want. The denominator is just the total probability of the evidence, computed by summing over all possible causes.
17. Bag 1 contains 3 red and 2 black balls; bag 2 contains 1 red and 4 black. A bag is chosen at random and a red ball is drawn. What is the probability it came from bag 1?
3/4. Here P(red | bag 1) = 3/5 and P(red | bag 2) = 1/5, with each bag equally likely at 1/2. The total probability of red is (1/2)(3/5) + (1/2)(1/5) = 3/10 + 1/10 = 2/5. Then P(bag 1 | red) = (3/10)/(2/5) = 3/4. Notice the answer is much larger than the prior 1/2, because red evidence favours the red-rich bag.
18. Why can you not simply reverse a conditional probability?
Because P(A|B) and P(B|A) have different denominators - P(B) and P(A) respectively. They are equal only when P(A) = P(B). In the previous question P(red | bag 1) is 3/5 while P(bag 1 | red) is 3/4, and treating them as interchangeable is exactly the error Bayes theorem exists to prevent.
19. What does the theorem of total probability say?
That the probability of an event equals the sum of its probabilities across a set of mutually exclusive and exhaustive causes, each weighted by the probability of that cause: P(B) = Σ P(B|Aⱼ)P(Aⱼ). It is the denominator of Bayes theorem, and it is how the 2/5 in question 17 was obtained.
20. Two cards are drawn without replacement. What is the probability both are kings?
1/221. The first card is a king with probability 4/52, and given that, the second is a king with probability 3/51 - the deck has changed. Multiplying gives (4/52)(3/51) = 12/2652 = 1/221. Using 4/52 twice would assume replacement and give a different, larger answer.
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Practice set 4: the binomial distribution
21. State the binomial probability formula.
P(X = r) = ⁿCr pʳ q^(n−r), where n is the number of trials, p the probability of success and q = 1 − p. The ⁿCr counts which trials were the successes, and the powers give the probability of any one such pattern. It is a counting problem multiplied by a probability, which is the chapter's recurring shape.
22. A fair coin is tossed 5 times. What is the probability of exactly 3 heads?
5/16. Using the formula with n = 5, r = 3, p = q = 1/2: ⁵C₃ (1/2)³(1/2)² = 10 × 1/32 = 10/32 = 5/16. The ⁵C₃ = 10 counts which three of the five tosses were heads, and each specific sequence has probability 1/32.
23. For 10 tosses of a fair coin, what are the mean and variance of the number of heads?
Mean 5 and variance 2.5. The mean is np = 10 × 0.5 = 5 and the variance is npq = 10 × 0.5 × 0.5 = 2.5. Note the variance is always smaller than the mean for a binomial distribution, since q is less than 1 - a useful check on any answer.
24. A fair die is rolled 4 times. What is the probability of at least one six?
671/1296, or about 0.518. Compute the complement: no six on any roll has probability (5/6)⁴ = 625/1296, so at least one six is 1 − 625/1296 = 671/1296. Adding the cases of exactly one, two, three and four sixes gives the same answer with four times the work.
25. What conditions must hold for a binomial model to apply?
A fixed number of trials, exactly two outcomes per trial, constant success probability, and independent trials. Sampling without replacement violates the last two, because each draw changes the composition of what remains - which is why question 20 could not be done binomially.
Practice set 5: mixed and conceptual
26. Why is the complement method preferred for 'at least one' questions?
Because 'at least one' spans many cases while 'none' is a single case. One minus P(none) gives the answer in one step, whereas summing the individual cases multiplies the work and the chance of omitting one. Question 24 is the standard illustration.
27. Why is drawing cards without replacement not a binomial experiment?
Because the trials are neither independent nor identically distributed. Removing a king changes the probability that the next card is a king, so p is not constant and the outcome of one draw affects the next. The correct model is the hypergeometric distribution, though JEE handles such cases by direct counting instead.
28. What is the expected value of a random variable?
The sum of each value multiplied by its probability - the long-run average. For a fair die it is (1 + 2 + 3 + 4 + 5 + 6)/6 = 3.5. Note the expected value need not be an attainable outcome; no single roll ever produces 3.5, which is a point questions like to make.
29. What is the difference between odds and probability?
Probability is favourable outcomes over total outcomes; odds in favour is favourable over unfavourable. If the probability is 1/4, the odds in favour are 1 to 3, not 1 to 4. Converting between them is a common short question, and the denominator is what differs.
30. A fair coin is tossed 3 times. What is the probability of at least one head?
7/8. The complement is no heads at all, which is TTT with probability (1/2)³ = 1/8, so the answer is 1 − 1/8 = 7/8. Enumerating gives the same result: seven of the eight equally likely sequences contain at least one head.
How to study this chapter efficiently
- ✓Secure Permutations and Combinations first. Most probability questions are counting problems with one division at the end.
- ✓Write the sample space size explicitly before computing anything. Most classical-probability errors are denominator errors.
- ✓For every 'at least' question, immediately consider the complement.
- ✓Keep mutually exclusive and independent in separate mental boxes, and remember that non-trivial events cannot be both.
- ✓For Bayes questions, write the prior, the likelihoods and the total probability as three labelled lines before combining them.
- ✓Check binomial applicability before using the formula - fixed n, two outcomes, constant p, independent trials.
Turn this into active practice
The conditional-probability errors in this chapter are stubborn because they feel right. Reversing P(A|B) and P(B|A) produces a plausible number and a confident wrong answer, and only being marked wrong makes the distinction stick.
The JEE Probability quiz on QUFF generates fresh questions across the basic rules, conditional probability, Bayes theorem and the binomial distribution, marks them instantly and explains each answer. Do a mixed set and note for each error whether the counting or the probability reasoning failed - they need different remedies.
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