An event and its complement partition the sample space
E and not E have no common outcome and together cover S. Therefore P(E)+P(not E)=1, often making indirect counting shorter.
E and not E have no common outcome and together cover S. Therefore P(E)+P(not E)=1, often making indirect counting shorter.
All elementary outcome probabilities sum to one; an event probability and its complement also sum to one, with each value from zero to one.
P(not E)=1-P(E). Subtract the event probability from one, not take its reciprocal, and verify the two values sum exactly to one.
Use complement when the excluded set is shorter. For A or B, form the union without double-counting their intersection; for A and B, keep only outcomes satisfying both.
For two coins, complement of at least one head is no head={TT}. Hence P(at least one head)=1-1/4=3/4; direct listing gives the same result.
If P(E):P(not E)=a:b, write probabilities as ak and bk. Since their sum is one, k=1/(a+b), giving a/(a+b) and b/(a+b).
List outcomes satisfying each number property. For or combine unique outcomes; for and retain only the intersection. Then divide by the unchanged total card count.
For weather, winning, defective items and success events, write the exact negated statement, subtract from one and add the pair back to verify.
Translate at least, at most, not, or, and and neither into explicit subsets; remove duplicates from unions and compare direct with complement counts.
Keep one sample denominator, list each favourable subset separately, note complement or overlap relationships, and check exhaustive disjoint parts sum to one.
Use known category probabilities to find a remaining probability, use a supplied count to recover total population and all category counts, then validate that counts and probabilities exhaust the sample space.
State the event relation, preserve one sample denominator, show direct or complement counts, answer each part and verify complement/exhaustive totals.
Solve complement, complement-ratio, overlapping-condition and multi-part population problems with exhaustive probability checks.