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3.9 · Paper 2B
Venn diagrams, tree diagrams, conditional probability and expected value — using probability to make fair decisions.
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A Venn diagram sorts outcomes into overlapping sets. $A \cap B$ means both, $A \cup B$ means either or both, and $A'$ means not $A$. The addition rule is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. For equally likely outcomes, probability = favourable ÷ total; Fermat's method lists all equally likely ways the rest of a game could go to share a prize fairly.
On a tree diagram you multiply along the branches and add the end results you want. Conditional probability: $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$. Events are independent if $P(A \cap B) = P(A)P(B)$ — knowing one happened tells you nothing about the other. Without replacement, the probabilities on later branches change.
The expected value is the long-run average: $E(X) = \sum x\,P(X = x)$. The expected frequency of an event is $n \times p$. A game is fair if the expected net winnings are 0. Use expected values to compare options — but remember they ignore how risky a one-off outcome is.
A game costs £2 to play. A fair dice is rolled: a 6 wins £10, anything else wins nothing. Is the game fair?
Net winnings $X$: a 6 gives $10 - 2 = +8$ with probability $\frac{1}{6}$; anything else gives $-2$ with probability $\frac{5}{6}$.
$E(X) = 8 \times \frac{1}{6} + (-2) \times \frac{5}{6} = \frac{8 - 10}{6} = -\frac{1}{3}$, about −33p per game. The game is not fair: in the long run the player loses about 33p each time.
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