1. Chapter at a Glance
A medical test is 99% accurate, yet a positive result may still be more likely to be a false alarm than a real disease — if the disease is rare. Understanding why needs conditional probability: the probability of an event given that another has happened. This chapter develops conditional probability, the multiplication rule, independent events, the theorem of total probability and Bayes' theorem (which "reverses" conditional probabilities), and then introduces random variables, their probability distributions and their mean (expected value).
- Conditional probability P(E|F) and its properties.
- Multiplication theorem P(E ∩ F) = P(F)P(E|F).
- Independent events.
- Partitions; theorem of total probability.
- Bayes' theorem and its applications.
- Random variables, probability distributions and the mean (expectation).
2. Conditional Probability
"The probability of E given that F has occurred." F becomes the new (reduced) sample space.
- P(F|F) = P(S|F) = 1.
- P(E′|F) = 1 − P(E|F).
- P((A ∪ B)|F) = P(A|F) + P(B|F) − P((A ∩ B)|F).
A family has two children. Find the probability that both are boys, given that at least one is a boy.
Solution: S = {BB, BG, GB, GG}. F = at least one boy = {BB, BG, GB}; E ∩ F = {BB}. P(E|F) = (1/4)/(3/4) = 1/3.
Two dice are thrown. Find P(sum = 8 | first die shows an even number).
Solution: F has 18 outcomes; E ∩ F = {(2,6), (4,4), (6,2)}. P = 3/18 = 1/6.
3. Multiplication Theorem and Independence
P(E ∩ F) = P(E) P(F|E) = P(F) P(E|F); for three events P(E ∩ F ∩ G) = P(E)P(F|E)P(G|E ∩ F).
An urn has 10 black and 5 white balls. Two balls are drawn one after the other without replacement. Find P(both black).
Solution: (10/15) × (9/14) = 3/7.
E and F are independent if P(E ∩ F) = P(E) P(F) (equivalently P(E|F) = P(E)). If E, F are independent, so are E and F′, E′ and F, E′ and F′.
Do not confuse with mutually exclusive (P(E ∩ F) = 0): two events with non-zero probabilities cannot be both.
P(A) = 0.3, P(B) = 0.4 and A, B are independent. Find P(A ∩ B), P(A ∪ B), P(neither).
Solution: 0.12; 0.3 + 0.4 − 0.12 = 0.58; neither = P(A′)P(B′) = 0.7 × 0.6 = 0.42.
4. Total Probability and Bayes' Theorem
Events E₁, E₂, …, Eₙ form a partition of S if they are pairwise disjoint, exhaustive and each has non-zero probability.
Total probability: P(A) = Σ P(Eⱼ) P(A|Eⱼ).
Bayes' theorem: P(Eᵢ|A) = P(Eᵢ) P(A|Eᵢ) / Σ P(Eⱼ) P(A|Eⱼ).
P(Eᵢ) are prior probabilities; P(Eᵢ|A) are posterior probabilities after observing A. (Named after Thomas Bayes, 18th century.)
Practice data: machines X, Y, Z produce 50%, 30% and 20% of the bolts; 3%, 4% and 5% of their output is defective. A bolt chosen at random is defective. Find the probability that it came from X.
Solution: P(D) = 0.5(0.03) + 0.3(0.04) + 0.2(0.05) = 0.015 + 0.012 + 0.010 = 0.037. P(X|D) = 0.015/0.037 = 15/37.
A disease affects 0.1% of people. A test detects it 99% of the time when present, but also gives a false positive in 0.5% of healthy people. If a person tests positive, what is the probability they have the disease?
Solution: P(+) = 0.001(0.99) + 0.999(0.005) = 0.00099 + 0.004995 = 0.005985. P(D|+) = 0.00099/0.005985 ≈ 0.165 — only about 16.5%, because the disease is rare.
5. Random Variables and Their Mean
A random variable X assigns a real number to each outcome. Its probability distribution lists the values xᵢ with probabilities pᵢ, where pᵢ ≥ 0 and Σpᵢ = 1.
Two coins are tossed; X = number of heads. Find the distribution and the mean.
Solution:
| X | 0 | 1 | 2 |
|---|---|---|---|
| P(X) | 1/4 | 1/2 | 1/4 |
E(X) = 0 + ½ + ½ = 1.
X takes values 0, 1, 2, 3 with P(X) = k, 2k, 3k, 4k. Find k and E(X).
Solution: 10k = 1 → k = 0.1. E(X) = 0 + 0.2 + 0.6 + 1.2 = 2.
6. Solved Problems (Board Pattern)
Bag I has 3 red and 4 black balls; Bag II has 5 red and 6 black. A bag is chosen at random and a ball drawn is red. Find P(it came from Bag II).
Solution: P(R|I) = 3/7, P(R|II) = 5/11. P(II|R) = (½ × 5/11)/(½ × 3/7 + ½ × 5/11) = (5/11)/(3/7 + 5/11) = (35/77)/(68/77) = 35/68.
A die is thrown. E = "multiple of 3", F = "even number". Are E and F independent?
Solution: P(E) = 1/3, P(F) = 1/2, P(E ∩ F) = P({6}) = 1/6 = P(E)P(F). Independent.
A game: roll a die; win ₹6 for a 6, ₹2 for a 4 or 5, and lose ₹1 otherwise. Find the expected gain.
Solution: E = 6(1/6) + 2(2/6) − 1(3/6) = 1 + 2/3 − 1/2 = ₹7/6 ≈ ₹1.17.
7. Master Formula Table
| Result | Formula |
|---|---|
| Conditional | P(E|F) = P(E ∩ F)/P(F) |
| Multiplication | P(E ∩ F) = P(F)P(E|F) |
| Independence | P(E ∩ F) = P(E)P(F) |
| Neither (independent) | P(E′)P(F′) |
| Total probability | P(A) = ΣP(Eⱼ)P(A|Eⱼ) |
| Bayes | P(Eᵢ|A) = P(Eᵢ)P(A|Eᵢ)/ΣP(Eⱼ)P(A|Eⱼ) |
| Distribution | pᵢ ≥ 0, Σpᵢ = 1 |
| Mean | E(X) = Σxᵢpᵢ |
8. Master Summary Table
| Situation | Tool |
|---|---|
| "Given that…" | Conditional probability |
| Successive draws | Multiplication theorem |
| Separate experiments (coins, two people) | Independence |
| Several sources of an outcome | Total probability |
| "Outcome observed; which source?" | Bayes' theorem |
| Numerical outcomes, average value | Random variable, E(X) |
9. Mind Map
- P(E|F)
- Reduced sample space
- P(E)P(F|E)
- Independent events
- Total probability
- Prior → posterior
- Distribution
- E(X) = Σxp
10. Tips and Tricks
- For Bayes problems, draw a tree: first branch = sources (E₁, E₂, …), second = outcome A.
- "At least one" with independent events: 1 − product of failure probabilities.
- Check that a probability distribution sums to 1 before using it.
- Confusing P(A|B) with P(B|A).
- Treating mutually exclusive events as independent.
- Using "with replacement" probabilities for "without replacement" draws.
11. Important One-liners
- P(E|F) uses F as the new sample space.
- Independent: P(E ∩ F) = P(E)P(F).
- Mutually exclusive events with non-zero probabilities are never independent.
- Bayes' theorem updates prior probabilities using evidence.
- The mean of a random variable is its long-run average value.
12. Exam-pattern Questions with Detailed Answers
If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, then P(A ∩ B) is:
- 0.32
- 0.25
- 0.1
- 0.4
Show answer
P(A)P(B|A) = 0.8 × 0.4 = 0.32.
Answer: (A)
If P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P(E|F) and P(F|E).
Show answer
P(E|F) = 0.2/0.3 = 2/3; P(F|E) = 0.2/0.6 = 1/3.
Answer: 2/3 and 1/3
A coin is tossed three times. Find the mean number of heads.
Show answer
X: 0, 1, 2, 3 with 1/8, 3/8, 3/8, 1/8. E = (0 + 3 + 6 + 3)/8 = 1.5.
Answer: 1.5
Two cards are drawn without replacement from a deck of 52. Find P(both are kings).
Show answer
(4/52)(3/51) = 1/221.
Answer: 1/221
A and B solve a problem independently with probabilities ½ and ⅓. Find P(the problem is solved).
Show answer
1 − (½)(⅔) = 1 − ⅓ = ⅔.
Answer: 2/3
A die is thrown. Given that the number is odd, find P(it is prime).
Show answer
Odd {1, 3, 5}; prime among them {3, 5}. 2/3.
Answer: 2/3
Is the following a probability distribution? X: 0, 1, 2 with P: 0.4, 0.4, 0.2. If yes, find E(X).
Show answer
All ≥ 0, sum 1 → yes. E = 0.4 + 0.4 = 0.8.
Answer: Yes; E(X) = 0.8
If A and B are independent with P(A) = 0.3 and P(B) = 0.6, find P(A and not B).
Show answer
0.3 × 0.4 = 0.12.
Answer: 0.12
Practice data: in a class, 60% are boys. 30% of boys and 20% of girls wear spectacles. A student wearing spectacles is chosen. Find P(the student is a girl).
Show answer
P(S) = 0.6(0.3) + 0.4(0.2) = 0.18 + 0.08 = 0.26. P(G|S) = 0.08/0.26 = 4/13.
Answer: 4/13
A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.
Show answer
P(six) = 1/6, P(report six | six) = 3/4; P(not six) = 5/6, P(report six | not six) = 1/4. P = (1/6)(3/4)/[(1/6)(3/4) + (5/6)(1/4)] = (3/24)/(8/24) = 3/8.
Answer: 3/8
Two numbers are selected at random without replacement from 1, 2, 3, 4, 5, 6. X = the larger of the two. Find E(X).
Show answer
15 pairs. P(X = k) = (k − 1)/15 for k = 2, …, 6. E = (2·1 + 3·2 + 4·3 + 5·4 + 6·5)/15 = 70/15 = 14/3.
Answer: 14/3
Show that if E and F are independent, then E and F′ are also independent.
Show answer
P(E ∩ F′) = P(E) − P(E ∩ F) = P(E) − P(E)P(F) = P(E)(1 − P(F)) = P(E)P(F′). ∎
Answer: Proved
13. Quick Revision Summary
- P(E|F) = P(E ∩ F)/P(F); multiplication P(E ∩ F) = P(F)P(E|F).
- Independent ⇔ P(E ∩ F) = P(E)P(F) (different from mutually exclusive).
- Total probability: P(A) = ΣP(Eⱼ)P(A|Eⱼ).
- Bayes: P(Eᵢ|A) = P(Eᵢ)P(A|Eᵢ)/P(A).
- Random variable distribution: pᵢ ≥ 0, Σpᵢ = 1; mean E(X) = Σxᵢpᵢ.