Meet Your Tutor
Probability becomes much clearer when events are named before numbers are substituted. I will help you translate wording into conditional probability, total probability and Bayes’ theorem, then test whether every answer lies between zero and one.
Probability is the one chapter in Class 12 Maths where the maths itself is gentle — you are mostly multiplying and adding fractions — and the reading is what trips people up. A question says “given that”, or “if it is known that”, or “find the probability that it came from the second bag”, and suddenly a student who can integrate by parts is staring at the page. So let us slow right down. In this chapter we will build one single idea — what happens to a probability when you are handed extra information — and then watch that one idea grow into conditional probability, the multiplication theorem, independence, total probability and finally Bayes’ theorem. Take it a section at a time. Nothing here is beyond you.
- A Quick Refresher: The Probability You Already Know
- Conditional Probability: What “Given That” Really Does
- Properties of Conditional Probability
- The Multiplication Theorem on Probability
- Independent Events (and Why They Are Not “Mutually Exclusive”)
- The Theorem of Total Probability
- Bayes’ Theorem: Working Backwards From the Evidence
- Choosing Between Total Probability and Bayes’
- Random Variables and Probability Distributions
- Mean (Expectation) of a Random Variable
- Board-Style Mixed Practice
Your Game Plan
- Spend ten minutes on the refresher. If P(A ∪ B) and complements feel shaky, everything later will feel shaky.
- Learn the conditional probability formula and then do nothing but “given that” questions until reading them stops feeling strange.
- Notice that the multiplication theorem is just the conditional formula rearranged. One formula, two faces.
- Separate independent from mutually exclusive once and for all. Examiners love this confusion.
- Draw a tree for every total-probability and Bayes’ question. Every single one. No exceptions.
- Finish with the worksheet, and write full reasoning — board marks in this chapter come from stated steps, not just the final fraction.
Study Notes
A Quick Refresher: The Probability You Already Know
Everything in this chapter sits on four facts you met in Class 11. Read them once, slowly, and check that each one still makes sense to you.
A random experiment is anything whose outcome you cannot predict — a coin toss, a card drawn from a shuffled pack. The sample space S is the set of all possible outcomes. An event is any subset of S. When all outcomes are equally likely,
P(E) = (number of outcomes favourable to E) ÷ (total number of outcomes in S)
2. P(E′) = 1 − P(E). The complement rule. Your best friend for “at least one” questions.
3. P(A ∪ B) = P(A) + P(B) − P(A ∩ B). The addition theorem — you subtract the overlap because you counted it twice.
4. If A and B are mutually exclusive (they cannot both happen), then P(A ∩ B) = 0, so P(A ∪ B) = P(A) + P(B).
Solution. Let C = “plays cricket”, F = “plays football”.
P(C) = 22/40, P(F) = 18/40, P(C ∩ F) = 9/40.
(i) P(C ∪ F) = 22/40 + 18/40 − 9/40 = 31/40 (= 0.775).
(ii) “Neither” is the complement of “at least one”, so P(C′ ∩ F′) = 1 − 31/40 = 9/40 (= 0.225).
Notice how much work fact 2 saved us in part (ii). Whenever you see “neither”, “none” or “at least one”, reach for the complement first.
Conditional Probability: What “Given That” Really Does
Here is the whole idea in one sentence: extra information shrinks the sample space.
Picture a die. Before you look, the sample space is {1, 2, 3, 4, 5, 6} and the probability of getting a 4 is 1/6. Now a friend peeks and tells you “it is an even number”. Nothing about the die has changed — but your world has. The only outcomes still alive are {2, 4, 6}. Out of those three, one is a 4. So now the probability of a 4 is 1/3.
That is conditional probability. We write it P(A | B), read “the probability of A given B”, and it means: among the outcomes where B happened, what fraction also have A?
P(A | B) = P(A ∩ B) ÷ P(B) The denominator is the event you were told about. The numerator is both things happening together. If P(B) = 0 the formula is undefined — you cannot be told that an impossible thing happened.
Why it works. If every outcome is equally likely, P(A ∩ B) = n(A ∩ B)/n(S) and P(B) = n(B)/n(S). Divide one by the other and the n(S) cancels, leaving n(A ∩ B)/n(B) — which is exactly “out of the B outcomes, how many are also A”. The formula is doing nothing more than re-measuring the world with B as the new whole.
Solution. A = {4, 5, 6} and B = {2, 4, 6}, so A ∩ B = {4, 6}.
P(A ∩ B) = 2/6, P(B) = 3/6.
P(A | B) = (2/6) ÷ (3/6) = 2/3.
Sanity check by counting: among the three even numbers, two of them (4 and 6) are greater than 3. Two out of three. The formula agrees with common sense — it always will.
Solution. Write the sample space in birth order: S = {BB, BG, GB, GG}, four equally likely outcomes.
Let A = “both are girls” = {GG}; B = “at least one girl” = {BG, GB, GG}.
A ∩ B = {GG}, so P(A ∩ B) = 1/4 and P(B) = 3/4.
P(A | B) = (1/4) ÷ (3/4) = 1/3.
Most people’s gut says 1/2. The gut is wrong, and this is exactly why the formula exists. “At least one girl” leaves three survivors, not two — BG and GB are genuinely different outcomes.
Solution. Every king is a face card, so K ∩ F = K and P(K ∩ F) = 4/52.
(i) P(K | F) = (4/52) ÷ (12/52) = 1/3. Among the 12 face cards, 4 are kings.
(ii) P(F | K) = (4/52) ÷ (4/52) = 1. If you already know it is a king, it is certainly a face card.
Keep this pair in mind: P(A | B) and P(B | A) are different questions with different answers. Mixing them up is the single most expensive mistake in this chapter.
Properties of Conditional Probability
Once you fix an event B and start asking P( · | B) for various events, something rather lovely happens: conditional probability behaves exactly like ordinary probability. All the old rules survive, with “| B” carried along for the ride. There are three properties you should be able to state and use.
P1. P(S | B) = 1 and P(B | B) = 1.
P2. If A and C are mutually exclusive, P((A ∪ C) | B) = P(A | B) + P(C | B).
More generally, P((A ∪ C) | B) = P(A | B) + P(C | B) − P((A ∩ C) | B).
P3. P(A′ | B) = 1 − P(A | B). The complement rule still works.
Why P3 works. Inside the shrunken world B, the outcomes split cleanly into “also A” and “not A”. Formally, A ∩ B and A′ ∩ B are mutually exclusive and together make up B, so P(A ∩ B) + P(A′ ∩ B) = P(B). Divide the whole line by P(B) and you get P(A | B) + P(A′ | B) = 1. Rearrange.
Solution.
(i) P(A | B) = 0.3 ÷ 0.5 = 3/5 = 0.6.
(ii) P(B | A) = 0.3 ÷ 0.6 = 1/2 = 0.5.
(iii) By P3, P(A′ | B) = 1 − 3/5 = 2/5 = 0.4.
(iv) P(A ∪ B) = 0.6 + 0.5 − 0.3 = 0.8.
Look at (i): P(A | B) = 0.6 = P(A). Knowing B told us nothing new about A. That is a preview of independence, which we meet shortly.
Solution. First the three individual conditionals:
P(A | C) = (1/5) ÷ (2/5) = 1/2.
P(B | C) = (3/20) ÷ (2/5) = (3/20) × (5/2) = 3/8.
P((A ∩ B) | C) = (1/20) ÷ (2/5) = (1/20) × (5/2) = 1/8.
Now apply P2 in its general form:
P((A ∪ B) | C) = 1/2 + 3/8 − 1/8 = 4/8 + 3/8 − 1/8 = 3/4 = 0.75.
That is the ordinary addition theorem, wearing a “| C” hat. Nothing new to memorise.
The Multiplication Theorem on Probability
Take the conditional probability formula and multiply both sides by P(B). That is it. That is the entire multiplication theorem. But look at what the rearranged form lets you do: it turns “find the probability that both happen” into a story you can walk through in order — first this, then that.
P(A ∩ B) = P(A) · P(B | A) = P(B) · P(A | B) For three events:
P(A ∩ B ∩ C) = P(A) · P(B | A) · P(C | A ∩ B) Read it left to right as a chain: probability of the first, times probability of the second knowing the first happened, times probability of the third knowing the first two happened.
This is the tool for every “drawn one after another without replacement” question. Without replacement means the second draw happens in a changed world — one ball fewer, and possibly one red fewer — which is precisely what a conditional probability describes.
Solution. Let R₁ = “first ball is red”, R₂ = “second ball is red”.
P(R₁) = 5/8 — there are 5 red out of 8 balls.
P(R₂ | R₁) = 4/7 — a red is gone, so 4 red remain out of 7 balls.
P(R₁ ∩ R₂) = (5/8) × (4/7) = 20/56 = 5/14 (≈ 0.357).
The two denominators, 8 then 7, are the giveaway that you are using the conditional version. If the ball had been put back, both would have been 8.
Solution. (i) P(W₁) = 4/10 and P(G₂ | W₁) = 6/9, so
P(W₁ ∩ G₂) = (4/10) × (6/9) = 24/90 = 4/15 (≈ 0.267).
(ii) “Both the same colour” splits into two mutually exclusive stories, so add them:
both white = (4/10) × (3/9) = 12/90;
both green = (6/10) × (5/9) = 30/90;
total = 42/90 = 7/15 (≈ 0.467).
Multiply along a story, add across different stories. Write that on the inside cover of your notebook.
Solution. Let K₁, K₂, K₃ be “king on the first / second / third draw”.
P(K₁) = 4/52; P(K₂ | K₁) = 3/51; P(K₃ | K₁ ∩ K₂) = 2/50.
P(K₁ ∩ K₂ ∩ K₃) = (4/52) × (3/51) × (2/50) = 24 ÷ 132600 = 1/5525 (≈ 0.000181).
Both the numerator and the denominator drop by one each time — one king used up, one card gone. If you can see that pattern, you can write a chain of any length.
Independent Events (and Why They Are Not “Mutually Exclusive”)
Sometimes the extra information is useless. You are told B happened, you recompute, and the probability of A does not budge. When that happens we say A and B are independent.
P(A ∩ B) = P(A) · P(B) Use the product form to test independence, because it needs no division and still works when a probability is 0.
Bonus fact: if A and B are independent, then so are A′ and B, A and B′, and A′ and B′.
Now the distinction that examiners return to year after year.
| Mutually exclusive | Independent | |
|---|---|---|
| Meaning | They cannot happen together | One happening tells you nothing about the other |
| Test | P(A ∩ B) = 0 | P(A ∩ B) = P(A)·P(B) |
| About | The sets — they do not overlap | The information — it does not help |
| Example | One die: “even” and “odd” | Two dice: “first is a 6” and “second is a 6” |
| Can both hold? | Only if P(A) = 0 or P(B) = 0. For real events with positive probability, mutually exclusive events are never independent. | |
Why that last row is true. If A and B are mutually exclusive then P(A ∩ B) = 0. For them to be independent we would also need P(A)·P(B) = 0, which forces one of them to be zero. Think about it in plain words: if you are told B happened, and A cannot happen alongside B, then P(A | B) = 0. That is about as far from “the information didn’t help” as you can get. Mutually exclusive events are the most informative about each other, not the least.
Solution. A = {2, 4, 6} so P(A) = 3/6 = 1/2. B = {3, 6} so P(B) = 2/6 = 1/3.
A ∩ B = {6}, so P(A ∩ B) = 1/6.
Test: P(A) · P(B) = (1/2)(1/3) = 1/6 = P(A ∩ B). ✓
So A and B are independent. They are not mutually exclusive, since the outcome 6 lies in both.
This single example is worth remembering, because it is a counter-example to the belief that “independent” and “mutually exclusive” are the same word twice.
Solution.
(i) P(A ∩ B) = (1/2)(1/3) = 1/6.
(ii) P(A ∪ B) = 1/2 + 1/3 − 1/6 = 3/6 + 2/6 − 1/6 = 2/3.
(iii) A′ and B′ are also independent, so P(A′ ∩ B′) = (1/2)(2/3) = 1/3.
Check: (ii) and (iii) must add to 1, because “at least one” and “neither” are complements. 2/3 + 1/3 = 1. ✓
Solution. Let A and B be “first solves it” and “second solves it”, with P(A) = 1/3, P(B) = 1/4.
(i) Go through the complement. P(neither solves) = P(A′)P(B′) = (2/3)(3/4) = 6/12 = 1/2.
So P(solved) = 1 − 1/2 = 1/2.
(ii) “Exactly one” means (A and not B) or (not A and B) — two mutually exclusive stories:
(1/3)(3/4) + (2/3)(1/4) = 3/12 + 2/12 = 5/12.
Cross-check: P(both) = (1/3)(1/4) = 1/12, and 5/12 + 1/12 = 6/12 = 1/2 = P(solved). ✓ The two ways of counting agree.
The Theorem of Total Probability
Some questions have a hidden first step. “A bag is chosen at random and then a ball is drawn” — before you can talk about the colour, you have to survive the choice of bag. The theorem of total probability is how we handle exactly this: split the situation into every possible first step, work out each branch, and add.
P(A) = P(E₁)·P(A | E₁) + P(E₂)·P(A | E₂) + … + P(En)·P(A | En) In words: for each branch, multiply the chance of taking that branch by the chance of A along it, then add all the branches.
Why it works. The pieces A ∩ E₁, A ∩ E₂, … do not overlap (because the E’s do not overlap) and they use up all of A (because the E’s cover everything). So their probabilities simply add to give P(A). Then each piece is rewritten with the multiplication theorem: P(A ∩ Ei) = P(Ei)·P(A | Ei). That is the whole proof, and it is worth being able to write it — it is a standard board question.
Solution. Partition by which bag was chosen: P(B₁) = P(B₂) = 1/2.
P(R | B₁) = 3/7 (3 red out of 7 balls). P(R | B₂) = 5/11 (5 red out of 11).
P(R) = (1/2)(3/7) + (1/2)(5/11) = 3/14 + 5/22.
Common denominator 154: 3/14 = 33/154 and 5/22 = 35/154.
P(R) = 68/154 = 34/77 (≈ 0.4416).
Sensible? Bag I is 42.9% red, Bag II is 45.5% red, so the answer must land between them. 44.2% does. Always run this check — it catches a surprising number of errors.
Solution. The three machines partition the output.
P(A) = 0.5, P(B) = 0.3, P(C) = 0.2 (these add to 1 — always check).
P(D | A) = 0.01, P(D | B) = 0.02, P(D | C) = 0.03.
P(D) = (0.5)(0.01) + (0.3)(0.02) + (0.2)(0.03)
= 0.005 + 0.006 + 0.006 = 0.017 = 17/1000.
So about 17 bolts in every thousand are defective. Notice that machine C, though it makes the least, contributes as much defective stock as machine B — that is the kind of remark that earns a full mark on an application question.
Solution. The hidden first step is the colour of the transferred ball. Partition on that.
Branch 1 — a red is transferred. Probability 3/8. Bag B now has 5 red and 4 black, 9 balls in all, so P(red | red transferred) = 5/9.
Branch 2 — a black is transferred. Probability 5/8. Bag B now has 4 red and 5 black, so P(red | black transferred) = 4/9.
P(red) = (3/8)(5/9) + (5/8)(4/9) = 15/72 + 20/72 = 35/72 (≈ 0.4861).
Answer: 35/72. Bag B was 50% red before the transfer and a slightly black-heavy bag fed into it, so a number a little under 1/2 is exactly what we should expect.
Bayes’ Theorem: Working Backwards From the Evidence
Total probability walks forwards: you know which machine, you want the chance of a defect. Bayes’ theorem walks backwards: you are holding a defective bolt, and you want to know which machine most likely made it. It is the maths of detective work — you observe an effect and reason back to the cause.
Let us build the picture first. Suppose Machine A makes 60% of a factory’s bolts and 3% of its bolts are defective, while Machine B makes the other 40% with a 5% defect rate. Here is the whole two-stage experiment on one tree.
P(Ei | A) = [ P(Ei)·P(A | Ei) ] ÷ [ Σj P(Ej)·P(A | Ej) ] Numerator = the one branch you are asked about. Denominator = all the branches added up. The denominator is just total probability, so Bayes’ theorem is total probability with one branch singled out.
Vocabulary: P(Ei) are the prior probabilities (before the evidence), P(A | Ei) are the likelihoods, and P(Ei | A) are the posterior probabilities (after the evidence).
Solution. The two “defective” leaves are all we need.
P(A ∩ D) = (3/5)(3/100) = 9/500.
P(B ∩ D) = (2/5)(1/20) = 1/50 = 10/500.
P(D) = 9/500 + 10/500 = 19/500 = 0.038.
P(A | D) = (9/500) ÷ (19/500) = 9/19 (≈ 0.4737), and so P(B | D) = 10/19 (≈ 0.5263).
Look what happened. Before seeing the bolt, Machine A was the more likely source (60%). After seeing that it is defective, Machine A drops to about 47% — the evidence pushed suspicion towards the sloppier machine. That shift is the entire point of Bayes’ theorem.
Solution. We already did the forward work in Example 13:
P(B₁ ∩ R) = (1/2)(3/7) = 3/14 = 33/154.
P(B₂ ∩ R) = (1/2)(5/11) = 5/22 = 35/154.
P(R) = 68/154 = 34/77.
P(B₁ | R) = (33/154) ÷ (68/154) = 33/68 (≈ 0.4853). Hence P(B₂ | R) = 35/68.
Once you have the leaves written down, a Bayes’ question costs you one extra line. That is why the tree is always worth drawing.
Solution. Let C = “has the condition”, T = “tests positive”.
P(C) = 1/200 = 0.005, P(C′) = 199/200 = 0.995.
P(T | C) = 99/100, P(T | C′) = 2/100.
Numerator: P(C)·P(T | C) = (1/200)(99/100) = 99/20000.
Other branch: P(C′)·P(T | C′) = (199/200)(2/100) = 398/20000.
P(T) = (99 + 398)/20000 = 497/20000.
P(C | T) = 99/497 ≈ 0.199, i.e. about 20%.
A test that is 99% accurate on sick people, and the positive result still only means a one-in-five chance of illness. The reason is that healthy people are so much more numerous that even a small 2% error rate produces four times as many false positives as there are true positives. This is why doctors retest. It is also the most quoted real-world use of Bayes’ theorem.
Solution. Let E = “the die actually showed a six”, so P(E) = 1/6 and P(E′) = 5/6. Let R = “he reports a six”.
If it really was a six, he reports a six when he tells the truth: P(R | E) = 4/5.
If it was not a six, he reports a six when he lies: P(R | E′) = 1/5.
P(E ∩ R) = (1/6)(4/5) = 4/30. P(E′ ∩ R) = (5/6)(1/5) = 5/30.
P(R) = 9/30 = 3/10.
P(E | R) = (4/30) ÷ (9/30) = 4/9 (≈ 0.444).
His word raised the probability of a six from 1/6 (≈ 0.167) to 4/9 (≈ 0.444) — strong evidence, but still not enough to make it likely. Do not fall into the trap of answering 4/5; that is his reliability, not the probability of the event.
Choosing Between Total Probability and Bayes’
In the exam hall the hard part is not the arithmetic — it is deciding which of the two theorems the question is asking for. Happily, the decision comes down to one question: which way round is time running?
• “…is drawn and is red. Find the probability that it came from…”
• “…tests positive. Find the probability that the person actually has…”
In each case something has already been observed, and you are asked to name the source. That is Bayes’, every time. And you still have to compute the total probability first — it becomes your denominator.
(i) Find the probability that this person has an accident this year.
(ii) Given that the person had an accident, find the probability that they are a scooter driver.
Solution. Total drivers = 3000 + 4000 + 5000 = 12000, so the priors are
P(S) = 3000/12000 = 1/4, P(C) = 4000/12000 = 1/3, P(T) = 5000/12000 = 5/12. (They add to 1. ✓)
(i) Total probability. With A = “has an accident”:
P(A) = (1/4)(1/100) + (1/3)(3/100) + (5/12)(15/100)
= 1/400 + 1/100 + 1/16
= 1/400 + 4/400 + 25/400 = 30/400 = 3/40 = 0.075.
(ii) Bayes’. P(S | A) = (1/400) ÷ (3/40) = (1/400) × (40/3) = 1/30 (≈ 0.0333).
For completeness, P(C | A) = (4/400) ÷ (3/40) = 2/15 and P(T | A) = (25/400) ÷ (3/40) = 5/6. Check: 1/30 + 2/15 + 5/6 = 1/30 + 4/30 + 25/30 = 1. ✓
Scooter drivers are a quarter of the customers but only a thirtieth of the accidents. Truck drivers are under half the customers and five-sixths of the accidents. Always check that your posterior probabilities add to 1 — it is the fastest error-catcher in the chapter.
Random Variables and Probability Distributions
Please confirm against the current syllabus copy issued by your own school before deciding how much time to give this section and the next one. It is short, it is useful, and it makes the rest of the chapter clearer — so reading it is never wasted. Just do not let it crowd out the five core topics above if your school’s copy leaves it out.
So far every event has been described in words — “the ball is red”, “the bolt is defective”. A random variable lets us describe outcomes with a number instead, which is far more convenient once you want to summarise or average things.
The probability distribution of X is the list of every value X can take together with its probability:
X : x₁, x₂, …, xn P(X) : p₁, p₂, …, pn It is a valid distribution only if every pi ≥ 0 and Σ pi = 1. Check that sum every single time — it is a free mark and a free error-check.
The everyday picture: toss two coins. The outcomes are HH, HT, TH, TT. Define X = the number of heads. Then X turns HH into 2, both HT and TH into 1, and TT into 0. Notice that X collapses several outcomes into one number — which is exactly why we have to add their probabilities together.
Solution. S = {HH, HT, TH, TT}, each of probability 1/4.
X = 0 only for TT → P(X = 0) = 1/4.
X = 1 for HT or TH → P(X = 1) = 1/4 + 1/4 = 1/2.
X = 2 only for HH → P(X = 2) = 1/4.
| X | 0 | 1 | 2 |
|---|---|---|---|
| P(X) | 1/4 | 1/2 | 1/4 |
Sum = 1/4 + 1/2 + 1/4 = 1. ✓
Solution. Total ways of choosing 2 bulbs from 6 = 6C₂ = 15.
P(X = 0) = 4C₂/15 = 6/15 = 2/5.
P(X = 1) = (4C₁ × 2C₁)/15 = 8/15.
P(X = 2) = 2C₂/15 = 1/15.
| X | 0 | 1 | 2 |
|---|---|---|---|
| P(X) | 6/15 | 8/15 | 1/15 |
Sum = (6 + 8 + 1)/15 = 15/15 = 1. ✓ Leaving the fractions over a common denominator of 15 makes that check instant — simplify only at the very end.
Solution. 52C₂ = 1326 equally likely pairs.
P(X = 0) = 48C₂/1326 = 1128/1326 = 188/221.
P(X = 1) = (4 × 48)/1326 = 192/1326 = 32/221.
P(X = 2) = 4C₂/1326 = 6/1326 = 1/221.
Check: (188 + 32 + 1)/221 = 221/221 = 1. ✓
Two aces in two cards is about a 0.45% chance — roughly once in 221 attempts. The numbers are ugly, but the method is identical to Example 22.
Mean (Expectation) of a Random Variable
You already know how to find the mean of a list of numbers: add them up and divide by how many there are. But what if some numbers are more likely than others? Then a plain average is unfair to the common values. The mean of a random variable fixes this by weighting each value by its probability.
μ = E(X) = Σ xi pi = x₁p₁ + x₂p₂ + … + xnpn It is the long-run average value of X if you repeated the experiment again and again. The mean need not be a value X can actually take — a fair die has mean 3.5, and no die face reads 3.5.
Why the formula looks like that. Imagine repeating the experiment 1000 times. You would expect the value x₁ to appear about 1000p₁ times, x₂ about 1000p₂ times, and so on. The ordinary average is then (1000p₁x₁ + 1000p₂x₂ + …)/1000, and the 1000s cancel, leaving Σ xipi. So the “weighted average” really is an ordinary average in disguise.
(ii) A fair die is rolled and Y is the number shown. Find E(Y).
Solution.
(i) From Example 21: E(X) = 0(1/4) + 1(1/2) + 2(1/4) = 0 + 1/2 + 1/2 = 1.
Comforting: on two coins you expect one head on average.
(ii) Each face has probability 1/6, so
E(Y) = (1 + 2 + 3 + 4 + 5 + 6)/6 = 21/6 = 7/2 = 3.5.
Here is that “impossible” mean — 3.5 is not a face of the die, but it is the correct long-run average.
Solution.
(i) E(X) = 0(6/15) + 1(8/15) + 2(1/15) = 0 + 8/15 + 2/15 = 10/15 = 2/3 (≈ 0.667).
Sensible: 2 of the 6 bulbs are defective, i.e. one third, and you drew 2 bulbs, so you expect 2 × (1/3) = 2/3 defectives. ✓
(ii) E(X) = 0(188/221) + 1(32/221) + 2(1/221) = (32 + 2)/221 = 34/221 = 2/13 (≈ 0.1538).
Sensible again: 4 aces out of 52 is 1/13, and you drew 2 cards, so 2 × (1/13) = 2/13. ✓
Both cross-checks used the same shortcut. It works nicely for “count how many of a type you drew” variables, and it is a wonderful way to catch an arithmetic slip — but write the full Σ xipi working in the exam, because that is what earns the marks.
Solution. P(X = 30) = 1/6, P(X = 10) = 2/6 = 1/3, P(X = −15) = 3/6 = 1/2. Sum = 1. ✓
E(X) = 30(1/6) + 10(1/3) + (−15)(1/2)
= 180/36 + 120/36 − 270/36 [common denominator 36]
= 30/36 = 5/6 ≈ ₹0.83 per game.
A positive expectation, so in the long run the player gains about 83 paise a game. If you were designing the stall you would want this number to be negative for the player. This is exactly how insurance premiums and lottery prizes are set — expectation is the working tool, not just an exam formula.
Board-Style Mixed Practice
Before the worksheet, let us pull the whole chapter together. Here is the decision table you should be able to reproduce from memory, followed by one full board-style answer written the way you should write yours.
| What the question sounds like | Tool to use | Formula |
|---|---|---|
| “…given that B has happened” | Conditional probability | P(A∩B) ÷ P(B) |
| “both”, “and then”, “without replacement” | Multiplication theorem | P(A)·P(B|A) |
| “independently”, “unrelated” | Independence | P(A)·P(B) |
| “at least one”, “none”, “neither” | Complement | 1 − P(none) |
| “a bag is chosen at random, then…” | Total probability | Σ P(Ei)·P(A|Ei) |
| “it is defective — find the probability it came from…” | Bayes’ theorem | one branch ÷ all branches |
| “find the probability distribution / mean of X” | Random variable | Σ pi = 1; μ = Σ xipi |
Step 1 — name the events.
E₁ = travels by bus, E₂ = travels by cycle, E₃ = walks. L = reaches late.
These three are mutually exclusive and cover every student, so they form a partition.
Step 2 — write the priors and check they add to 1.
P(E₁) = 0.40, P(E₂) = 0.35, P(E₃) = 0.25. Sum = 1. ✓
Step 3 — write the likelihoods.
P(L | E₁) = 0.05, P(L | E₂) = 0.10, P(L | E₃) = 0.15.
Step 4 — total probability for the denominator.
P(L) = (0.40)(0.05) + (0.35)(0.10) + (0.25)(0.15)
= 0.0200 + 0.0350 + 0.0375 = 0.0925 = 37/400.
Step 5 — apply Bayes’ theorem.
P(E₃ | L) = 0.0375 ÷ 0.0925 = 375/925 = 15/37 ≈ 0.405.
Step 6 — check. P(E₁ | L) = 200/925 = 8/37 and P(E₂ | L) = 350/925 = 14/37. Then 8/37 + 14/37 + 15/37 = 1. ✓
Walkers are only a quarter of the school but produce about 41% of the late arrivals. Notice that the answer is a clean fraction — if yours is not, suspect an arithmetic slip and redo the denominator.
Practice Worksheet
Work these on paper first, with full steps, before you open any answer. If a question defeats you, go back to the section it came from rather than reading the solution — that is where the learning is.
Q1. A card is drawn from a well-shuffled pack of 52. Given that the card is a spade, find the probability that it is a face card.
Show Answer
P(F ∩ S) = 3/52, P(S) = 13/52.
P(F | S) = (3/52) ÷ (13/52) = 3/13 ≈ 0.231.
Q2. Two dice are thrown together. Find the probability that the sum is 7, given that at least one die shows a 3.
Show Answer
Sum 7 outcomes: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). Of these, only (3,4) and (4,3) contain a 3, so P(A ∩ B) = 2/36.
P(A | B) = (2/36) ÷ (11/36) = 2/11 ≈ 0.182.
Watch out: the denominator is 11, not 12 — (3,3) must not be double-counted.
Q3. A box has 7 red and 5 green balls. Two balls are drawn one after another without replacement. Find (i) P(both green) and (ii) P(second is red, given the first was green).
Show Answer
(i) P(G₁ ∩ G₂) = (5/12) × (4/11) = 20/132 = 5/33 ≈ 0.152.
(ii) After a green is removed, 11 balls remain of which 7 are still red, so P(R₂ | G₁) = 7/11 ≈ 0.636.
Q4. A and B are independent events with P(A) = 0.4 and P(B) = 0.7. Find (i) P(A ∩ B), (ii) P(A ∪ B), (iii) P(A′ ∩ B), (iv) P(A | B).
Show Answer
(ii) P(A ∪ B) = 0.4 + 0.7 − 0.28 = 0.82 = 41/50.
(iii) P(A′ ∩ B) = P(B) − P(A ∩ B) = 0.7 − 0.28 = 0.42 = 21/50. (Or (0.6)(0.7) = 0.42, since A′ and B are independent too.)
(iv) Because they are independent, P(A | B) = P(A) = 0.4 = 2/5. No calculation needed — that is what independence means.
Q5. Two archers shoot at a target independently. Their probabilities of hitting it are 3/5 and 2/3. Find the probability that (i) the target is hit, (ii) exactly one archer hits it.
Show Answer
(ii) P(exactly one) = (3/5)(1/3) + (2/5)(2/3) = 3/15 + 4/15 = 7/15 ≈ 0.467.
Check: P(both hit) = (3/5)(2/3) = 6/15, and 7/15 + 6/15 = 13/15 = P(hit). ✓
Q6. Bag P contains 4 white and 5 black balls; Bag Q contains 6 white and 3 black balls. A bag is chosen at random and one ball is drawn. Find the probability that the ball is white.
Show Answer
P(W | P) = 4/9 and P(W | Q) = 6/9.
P(W) = (1/2)(4/9) + (1/2)(6/9) = 2/9 + 3/9 = 5/9 ≈ 0.556.
Sensible: it is the plain average of 4/9 and 6/9, because the two bags are equally likely and equally sized.
Q7. In Q6 the ball drawn turns out to be white. Find the probability that it came from Bag Q.
Show Answer
P(Q | W) = [(1/2)(6/9)] ÷ (5/9) = (3/9) ÷ (5/9) = 3/5 = 0.6.
So P(P | W) = 2/5, and the two add to 1. ✓ The white ball makes Bag Q the more likely source, which matches the fact that Bag Q is the whiter bag.
Q8. A company runs three plants X, Y and Z which produce 25%, 35% and 40% of its output. Of their output, 5%, 4% and 2% respectively are defective. An item is chosen at random from the total output. (i) Find the probability that it is defective. (ii) Given that it is defective, find the probability that it came from plant Y.
Show Answer
Likelihoods: P(D | X) = 0.05, P(D | Y) = 0.04, P(D | Z) = 0.02.
(i) P(D) = (0.25)(0.05) + (0.35)(0.04) + (0.40)(0.02)
= 0.0125 + 0.0140 + 0.0080 = 0.0345 = 69/2000.
(ii) P(Y | D) = 0.0140 ÷ 0.0345 = 140/345 = 28/69 ≈ 0.406.
Check: P(X | D) = 25/69 and P(Z | D) = 16/69; 25 + 28 + 16 = 69. ✓
Q9. A speaks the truth 3 times out of 4 and B speaks the truth 4 times out of 5. They independently describe the same incident. Find the probability that they contradict each other.
Show Answer
P(A truthful, B lying) = (3/4)(1/5) = 3/20.
P(A lying, B truthful) = (1/4)(4/5) = 4/20.
These two cases are mutually exclusive, so P(contradict) = 3/20 + 4/20 = 7/20 = 0.35.
Note: “both lie” is agreement, not contradiction, in this standard reading — if they both give the same false account they do not contradict each other.
Q10. A random variable X takes the values 0, 1, 2 and 3 with probabilities k, 2k, 3k and 4k. Find (i) k, (ii) P(X ≥ 2), (iii) the mean of X.
Show Answer
The distribution is 1/10, 2/10, 3/10, 4/10.
(ii) P(X ≥ 2) = P(X = 2) + P(X = 3) = 3/10 + 4/10 = 7/10 = 0.7.
(iii) μ = 0(1/10) + 1(2/10) + 2(3/10) + 3(4/10) = (0 + 2 + 6 + 12)/10 = 20/10 = 2.
Q11. Three fair coins are tossed. Let X be the number of tails. Write the probability distribution of X and find its mean.
Show Answer
X = 0: 1 outcome → 1/8. X = 1: 3 outcomes → 3/8. X = 2: 3 outcomes → 3/8. X = 3: 1 outcome → 1/8.
Sum = (1 + 3 + 3 + 1)/8 = 1. ✓
μ = 0(1/8) + 1(3/8) + 2(3/8) + 3(1/8) = (0 + 3 + 6 + 3)/8 = 12/8 = 3/2 = 1.5.
Sensible: three coins, each tail with probability 1/2, so on average 1.5 tails.
Q12. A bag contains 3 red and 5 white balls. A ball is drawn at random and, without replacing it, a second ball is drawn. Find the probability that (i) the second ball is red, (ii) both balls are of different colours.
Show Answer
(i) By total probability,
P(R₂) = (3/8)(2/7) + (5/8)(3/7) = 6/56 + 15/56 = 21/56 = 3/8 = 0.375.
A lovely result: the second ball is red with the same probability as the first. Before you look at anything, the second position is just as likely to hold a red as the first — the balls do not know which order they were drawn in.
(ii) Different colours means red-then-white or white-then-red:
(3/8)(5/7) + (5/8)(3/7) = 15/56 + 15/56 = 30/56 = 15/28 ≈ 0.536.
Kaizen: aim for one more correct question than yesterday. That’s it. One more.