1Count the total firstP(E)=totalfavourabletwo dice: 36 outcomes; three coins: 8
2Sums on two diceways to roll a sum of s=6−∣s−7∣s=2 to 12;a sum of 7 has the most ways, 6
3Flip to the complementP(at least one)=1−P(none)
4Either or bothP(A∪B)=P(A)+P(B)−P(A∩B)neither=1−P(A∪B);exclusive: add
5Independent eventsP(A∩B)=P(A)×P(B)all must happen: multiply
6Without replacementDraw one after another and shrink the fractions: two aces from a pack is 524×513, not 524×524.
7Count selections both waysWhen order does not matter, count the favourable cases and the total both as selections nCr; mixing ordered and unordered counts is the classic slip.
8Exactly r successesnCrpr(1−p)n−rn tries;p=chance of success on one try
9Taking turnsP(first wins)=1−(1−p)(1−q)pp=first player’s chance per turn,q=second’s
10Conditional probabilityP(A∣B)=P(B)P(A∩B)shrink the total to the outcomes where B happened
11Which source? (Bayes)When the result is known and the source is asked, find each route's chance (source, then result) and divide the route you want by the sum of all routes.
Test yourself
10 questions · 3 easy · 4 medium · 3 hard
Tap an option to answer. The first tap counts.
Q1Easy
At a fest game stall, you roll two fair dice and win if the total is exactly 10. What is the probability of winning?
Correct answer·121
Why
Count the total first: two dice give 6×6=36 equally likely outcomes.
A total of 10 comes from (4, 6), (5, 5) and (6, 4), which is 3 outcomes.
So the probability is 363=121.
Q2Easy
A shooter hits a target with probability 31 on each shot, and the shots are independent. If she fires 3 shots, what is the probability that she hits the target at least once?
Correct answer·2719
Why
At least one? Flip it: find the chance that every shot misses.
All three miss: (32)3=278.
So at least one hit has probability 1−278=2719; adding 31 three times wrongly gives 1.
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