conditional probability
Imagine you are told a die came up even. Now what is the chance it was a 6? Before the news, every face had a one-in-six chance. But hearing 'it was even' rules out 1, 3, and 5, leaving only 2, 4, 6 as live possibilities. Among those three, exactly one is a 6, so the chance jumps to one-in-three. Conditional probability is the machinery for exactly this kind of update: it answers 'given that B happened, how likely is A?', written P(A given B).
The definition is a ratio. To get P(A given B), restrict your attention to the worlds where B happened, then ask what fraction of THOSE also have A. In symbols, P(A given B) = P(A and B) / P(B), defined whenever P(B) is not zero. The denominator P(B) renormalizes: it scales the leftover probability so that the new, shrunken world again sums to 1. In the die example, P(A and B) = P(rolling a 6) = 1/6 and P(B) = P(even) = 1/2, so P(6 given even) = (1/6) / (1/2) = 1/3, matching our count.
Conditioning is how probability handles information. Every time you learn something, you can recompute the odds of everything else in light of it. A crucial honesty point: P(A given B) and P(B given A) are NOT the same number, and confusing them is the root of many famous blunders (the prosecutor's fallacy, the base-rate fallacy). The probability of a positive test given disease is not the probability of disease given a positive test.
Two fair coins are tossed. Given that at least one is heads, what is the probability both are heads? The equally likely outcomes are HH, HT, TH, TT. 'At least one heads' leaves HH, HT, TH — three worlds. Only HH has both heads, so the answer is 1/3, not 1/2.
Conditioning on 'at least one head' shrinks four worlds to three, and the answer is 1/3.
P(A given B) is undefined when P(B) = 0; you cannot condition on something that never happens (the measure-theoretic version handles conditioning on probability-zero events differently).