Conditional Probability

Use the definition $P(A|B) =\frac{ P(A\cap{}B)}{P}(B)$: divide the probability that both events occur together by the probability of the condition. Die example: B = "even number" has $P(B) =\frac{ 1}{2}$, A = "number greater than 3". Jointly favourable are {4, 6}, so $P(A\cap{}B) =\frac{ 2}{6} =\frac{ 1}{3}$. Hence $P(A|B) =\frac{ (\frac{1}{3})}{(\frac{1}{2})} =\frac{ 2}{3}$. Intuitively the sample space shrinks to the three even numbers {2, 4, 6}, of which two exceed 3. Depending on the task you find $P(A\cap{}B)$ in the contingency table (inner cell), in the tree diagram (path product) or by direct counting. The decisive step is to identify condition and target event cleanly: what is given, what is asked?

The direction of conditioning is swapped, and that usually changes the value drastically. P(A|B) asks: how likely is A given that B has occurred? P(B|A) asks the reverse. Famous medical test example: P(positive|ill) is the sensitivity of the test, often 99%. P(ill|positive) is what the tested person wants to know, and can still be small for a rare disease, because among the positives are many false alarms from the large healthy majority. The two quantities also have different denominators: P(A|B) normalizes to B, P(B|A) to A. They can only be converted via Bayes theorem: $P(B|A) =\frac{ P(A|B)\cdot{}P(B)}{P}(A)$. Carelessly equating the two directions is called the prosecutor's fallacy and is one of the most consequential statistical errors.

Both representations organize the same information, but with different strengths. The contingency table holds the joint probabilities $P(A\cap{}B)$ etc. in its four inner cells and the marginal probabilities at the edges. You read off a conditional probability by dividing an inner cell by its marginal total: P(A|B) = cell(A∩B)/margin(B). The tree diagram carries conditional probabilities directly on the second-level branches; the path rule $P(A\cap{}B) = P(B)\cdot{}P(A|B)$ multiplies along the path. Rule of thumb: if shares and percentages of a fixed population are given, the table is faster; if the task describes a temporal process in stages, the tree is more natural. For "reversed" questions (Bayes) it often helps to redraw the tree with the stages swapped.

When the occurrence of one does not change the probability of the other: $P(A|B) = P(A)$. Equivalent and computationally handier is the product formula $P(A\cap{}B) = P(A)\cdot{}P(B)$; this is exactly what you verify in tasks. Example: in a double coin toss, "first coin heads" and "second coin heads" are independent: $P(A\cap{}B) =\frac{ 1}{4} = (\frac{1}{2})\cdot{}(\frac{1}{2})$ ✓. Counterexample die: A = "number > 3" and B = "even" are dependent, since $P(A\cap{}B) =\frac{ 1}{3}$, but $P(A)\cdot{}P(B) = (\frac{1}{2})\cdot{}(\frac{1}{2}) =\frac{ 1}{4}$. Important: independence is not mutual exclusivity! Disjoint events with positive probabilities are maximally dependent, because one rules out the other. And real-world independence should be justifiable, for example by separate random mechanisms.

Bayes theorem is a direct consequence of the definition. The joint probability can be decomposed in two ways: $P(A\cap{}B) = P(A|B)\cdot{}P(B) = P(B|A)\cdot{}P(A)$. Equating both expressions and solving for P(B|A) yields Bayes theorem: $P(B|A) =\frac{ P(A|B)\cdot{}P(B)}{P}(A)$. It answers exactly the question of how to convert one conditioning direction into the other. The denominator P(A) is usually determined via total probability: P(A) = P(A|B)·P(B) + P(A|B̄)·P(B̄), the two tree-diagram paths leading to A. Typical exam application: from the sensitivity and false-positive rate of a test, compute the probability of really being ill given a positive result.