Along a branch (AND) $P(A)\times P(B)$multiply the probabilities ALONG the branches
Between routes (OR) $P(\text{route 1})+P(\text{route 2})$add the separate routes that both count
Without replacement $\dfrac{g}{n}\times\dfrac{g-1}{n-1}$the second pick has one fewer on top AND bottom
At least one $P(\ge 1)=1-P(\text{none})$the complement is almost always quicker
Both (with replacement) · $3$ red of $5$, replaced
P(both red) $=\dfrac{3}{5}\times\dfrac{3}{5}$
$=\dfrac{9}{25}$
Both (without replacement) · $3$ red of $5$, not replaced
P(both red) $=\dfrac{3}{5}\times\dfrac{2}{4}$
$=\dfrac{6}{20}=\dfrac{3}{10}$
One of each · $3$ red, $2$ blue, replaced
red-blue OR blue-red
$\dfrac{3}{5}\times\dfrac{2}{5}+\dfrac{2}{5}\times\dfrac{3}{5}=\dfrac{12}{25}$
Independent: one event does not affect the other (e.g. with replacement).
Without replacement: the item is not put back, so the second probability changes.
Outcome: a result of one stage of the experiment, e.g. "red".
Complement: "not happening": $P(\text{not }A)=1-P(A)$.
Branch: one path on the tree; probabilities along a branch multiply.
✗ Using the same fraction twice without replacement
✓ the second pick has one fewer item on the top and the bottom.
✗ Adding along a branch
✓ multiply along a branch (AND); only add between separate routes (OR).
✗ Branch pair not summing to $1$
✓ the two branches from a point always add to $1$ – check it.
✗ Only one route for "one of each"
✓ red-then-blue and blue-then-red – add both routes.
• The two branches from any point add up to $1$ – use that to fill a missing branch.
• Multiply ALONG branches (AND); add BETWEEN separate routes (OR).
• For "at least one", it is usually quicker to work out $1-P(\text{none})$.
• Label every branch with its probability before you calculate anything.
• AND $\to \times$ (along); OR $\to +$ (between routes).
• "At least one" $=1-P(\text{none})$.
mathedup.co.uk · sheet U87J