Learn first: Basics of combinatorics
Probability measures how "expected" a random event is with a number from 0 to 1. Exam problems mostly involve the classical definition, counting with combinatorics, adding and multiplying events, and "at least one" questions. The key to success is first picturing clearly all the equally likely outcomes.
P(A) = m/n — m favourable outcomes, n all equally likely outcomes; 0 ≤ P(A) ≤ 1
The crucial condition is that the outcomes are equally likely. When two dice are thrown, the sums (from 2 to 12) are not equally likely, but the pairs (1; 1), (1; 2), …, (6; 6) are — there are 36 of them.
Two dice are thrown. What is the probability that the total is 8?
When there are many outcomes, we count them with combinations Cₙᵏ instead of listing them.
3 of 10 lottery tickets are winning. Two tickets are taken at random. Find the probability that: a) both win; b) exactly one wins.
Mutually exclusive: P(A + B) = P(A) + P(B)
In general: P(A + B) = P(A) + P(B) − P(AB)
Why subtract P(AB)? Outcomes belonging to both events are counted in P(A) and in P(B) — twice.
One card is drawn from a 36-card deck. What is the probability that it is an ace or a heart?
If one event happening does not change the probability of the other, the events are independent:
P(AB) = P(A)·P(B)
Otherwise we use conditional probability:
P(AB) = P(A)·P(B | A)
The typical dependence is drawing without replacement: the first ball drawn changes what is left.
"At least one" — via the complement: P = 1 − P(none).
Two shooters fire once each; their hit probabilities are 0.8 and 0.7. Find the probability that: a) both hit; b) at least one hits; c) exactly one hits.
A bag holds 8 blue and 8 red balls. Two are drawn one after another without replacement. What is the probability that both are blue?
Bernoulli's formula: an experiment is repeated independently n times, with "success" probability p each time and q = 1 − p:
Pₙ(k) = Cₙᵏ pᵏ qⁿ⁻ᵏ
For example, in 4 coin tosses exactly 2 heads appear with probability C₄²·(1/2)⁴ = 6/16 = 3/8.
Geometric probability: if a point is thrown at random into a region, the probability is a ratio of measures (lengths, areas):
P = S(favourable region)/S(whole region)
For example, a point thrown into a square of side 2 lands in the inscribed circle with probability π·1²/4 = π/4.
Conditional probability P(B | A) is the probability of B given that A has happened. To compute it, we replace the whole sample space by the outcomes favourable to A:
P(B | A) = P(AB)/P(A), where P(A) > 0
For example, a die is rolled and we are told the result is even. The probability that it is greater than 5: the even outcomes are 2, 4, 6, and only 6 is greater than 5, so the conditional probability is 1/3, while the unconditional one is only 1/6. The condition changes the probability because it narrows the set of possible outcomes.
In many problems the result can come about in several ‘ways’: an item comes from different factories, a ball is drawn from different boxes. If H₁, H₂, …, Hₙ are mutually exclusive hypotheses forming a complete group (their probabilities add up to 1), then:
P(A) = P(H₁)·P(A | H₁) + P(H₂)·P(A | H₂) + … + P(Hₙ)·P(A | Hₙ)
This is the law of total probability. It is convenient to picture it as a tree: probabilities along one branch are multiplied, and the branches leading to the desired result are added.
The reverse question — given that A happened, how likely is it that it happened via a particular hypothesis — is answered by Bayes’ formula: P(Hᵢ | A) = P(Hᵢ)·P(A | Hᵢ)/P(A).
60% of the bulbs in a shop come from the first factory and 40% from the second. 2% of the first factory’s output and 5% of the second’s is faulty. Find the probability that a randomly chosen bulb is faulty, and then the probability that a faulty bulb came from the first factory.