Mathematics RU

Computing probabilities

Probabilities: equally likely outcomes, “at least one” through the complement, independent events, total probability and Bayes' formula on a medical test.

How to solve it

If all outcomes are equally likely, a probability is the share of favourable outcomes. Complex events break into simple ones: “and” for independent events is a product, “or” for exclusive ones a sum, and “at least one” is easiest through its opposite, “none”.

Step by step

Equally likely outcomes: count all outcomes and the favourable ones; the probability is their ratio. Two dice give $36$ ordered pairs.
“At least one”: $P = 1 - P(\text{none})$.
Independent events: the probability that all happen is the product of the probabilities.
Bayes: imagine a large group (say $10\,000$ people), count how many land in each branch, and take the share you need.

The hypothesis: say, “the person is ill”.

The observation: “the test is positive”.

The total probability of the observation over all hypotheses: the ill with a positive test plus the healthy with a false alarm.

Example: $2\%$ are ill, the test finds $95\%$ of them and errs on $10\%$ of the healthy. Out of $10\,000$: $190$ ill positives and $980$ healthy positives; $P = \frac{190}{1170} = \frac{19}{117} \approx 16\%$.

Common mistakes

Confusing $P(A \mid H)$ with $P(H \mid A)$: “the test finds $95\%$ of the ill” does not mean a positive test is $95\%$ illness.
Computing “at least one six in $4$ throws” as $\frac{4}{6}$ — probabilities do not add that way.
Treating the outcomes of two dice as unordered and getting $21$ instead of $36$.
Multiplying the probabilities of dependent events without the condition.