5.3 Probability

For repeated identical items, divide by factorials for duplicate arrangements; for selections with repetition, distinguish stars-and-bars style counting from ordinary combinations.

Write a small case first to check what an outcome means, then generalise. Position restrictions can change the formula entirely.

The distinct arrangements of A,A,B,C are 4!/2!, because swapping the A’s creates no new arrangement.

Treating identical objects as labelled creates artificial outcomes and inflates the count.

P(A∪B)=P(A)+P(B)−P(A∩B), P(Aᶜ)=1−P(A), and P(A|B)=P(A∩B)/P(B) when P(B)>0.

Draw a Venn diagram or tree when events overlap, and identify the denominator in conditional probability before substituting numbers.

If P(A)=0.6, P(B)=0.5 and P(A∩B)=0.2, then P(A∪B)=0.9.

Adding P(A) and P(B) without subtracting overlap can exceed 1.

Events are mutually exclusive when A∩B=∅. They are independent when P(A∩B)=P(A)P(B), equivalently P(A|B)=P(A) when defined.

Disjoint non-zero events cannot be independent: learning that one occurred makes the other impossible. Test the stated relationship with the correct equation.

A single die roll being even and odd is mutually exclusive; two independent coin tosses are not mutually exclusive across different tosses.

“Independent” does not mean unrelated in everyday language, and mutually exclusive events are not independent unless one has probability zero.

For equally likely outcomes, P(A)=number favourable/number total. For unequal outcomes, use the given probabilities and ensure they sum to one.

Define the sample space, avoid counting outcomes with different probabilities as equal, and use complements when the direct event is cumbersome.

The probability of at least one six in two fair rolls is 1−(5/6)²=11/36.

“At least one” includes two occurrences; subtracting only the probability of exactly one misses cases.