Use P(A and B) = P(A) × P(B) for independent events, and check whether learning one event happened changes the other's probability.
At the prize table, 'or' counted either description. Now two stalls ask for 'and': winning both. In this model the results are independent. P(win coconuts) = 0.2 and P(win tins) = 0.25. Winning both cannot be more likely than winning either game alone. Adding gives 0.45, which is above both chances, so addition cannot answer this 'and' question.
In our chance model, 0.2 of 100 parts is 20. Independence means 0.25 of those 20 parts, or 5, also wins the tins. So P(both) = 5/100 = 0.05 = 0.2 × 0.25. The rule is P(A and B) = P(A) × P(B) for independent events. For three mutually independent events, multiply all three probabilities. Each independent fair coin flip has a 0.5 chance of heads. To find P(B), divide P(A and B) by P(A), but only when independence is given and P(A) > 0.
Independent means learning one event happened leaves the other's probability unchanged. Different equipment alone does not prove this; a player's skill could link two games. With all three probabilities, test P(A and B) = P(A) × P(B). Mutually exclusive events with positive probabilities are dependent: learning one happened makes the other's chance 0. If either probability is 0, the product test holds and the events are independent.
A drum holds 20 tickets, 5 winners. Each draw picks uniformly at random. After a winning ticket is kept, 19 tickets remain with 4 winners: the next chance is 4/19, not 5/20. The draws are dependent. For two wins, multiply the first chance by the updated chance: (5/20) × (4/19), not 0.25 × 0.25. Replacing the ticket and mixing restores the original chance, making the draws independent.
Removing a winning ticket changes both the number of winners and the total.
For independent events, 'and' means multiply. A probability is between 0 and 1, so multiplying by it cannot increase a value. The chance of both is at most either individual chance; 'or' gives at least either one. Equality is possible. For 'or', add and subtract overlap. Loss has chance 1 - P(win). Independent wins give independent losses. Multiply for neither; subtract that result from 1 for at least one. Earlier independent fair coin flips leave the next heads chance at 0.5.
For independent events, P(A and B) = P(A) × P(B). Check whether learning one result changes the other's chance. In our raffle without replacement, update the second winning chance.
The rewritten readings were written for this dataset. Other readings adapt Khan Academy's 'Conditional probability and independence'. Khan Academy videos 'Compound probability of independent events' and 'Conditional probability and independence' are linked, not adapted. 'Single chances as building blocks' and nine practice questions were written for os-taxonomy. Source-version license terms are unknown.
You used 'or' at the prize table. Now two stalls ask for 'and'. In this model, the results are independent. P(win coconuts) = 0.2 and P(win tins) = 0.25. Leo adds them to get 0.45 for winning both.
Without doing any arithmetic, why can 0.45 not be right?
Jobs that lean on this skill. Follow one to see everything it is built on.
33 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.