Without replacement, update both the total remaining and the count of each colour before writing the next branches. The second probabilities depend on the first result, but multiplying probabilities along a path still gives that path’s probability. Example Draw twice without replacement from three red and two blue counters. After a red, two red and two blue remain. After a blue, three red and one blue remain. 35 -> R -> 25 -> B -> 24 -> R -> 24 -> B -> 34 -> R -> 14 -> B P(RR)= 35 x 24= 310, whereas P(BB)= 25 x 14= 110. P(different colours)= 35 x 24+ 25 x 34 = 310+ 310= 35. Check using the complement: 1- 310- 110= 35. Every second-stage denominator is 4; the red numerator is 2 after red but 3 after blue. If counters are added, removed or returned according to a stated rule, redraw the composition at each junction first. Do not reuse the first probability merely because the colour name is unchanged.

Without replacement, update both the total remaining and the count of each colour before writing the next branches. The second probabilities depend on the first result, but multiplying probabilities along a path still gives that path’s probability.

Draw twice without replacement from three red and two blue counters. After a red, two red and two blue remain. After a blue, three red and one blue remain.

\(\mathrm P(RR)=\dfrac35\times\dfrac24=\dfrac3{10}\), whereas \(\mathrm P(BB)=\dfrac25\times\dfrac14=\dfrac1{10}\).

\(\begin{aligned}\mathrm P(\text{different colours})&=\dfrac35\times\dfrac24+\dfrac25\times\dfrac34\\&=\dfrac3{10}+\dfrac3{10}=\dfrac35.\end{aligned}\)

Check using the complement: \(1-\dfrac3{10}-\dfrac1{10}=\dfrac35\). Every second-stage denominator is \(4\); the red numerator is \(2\) after red but \(3\) after blue.

If counters are added, removed or returned according to a stated rule, redraw the composition at each junction first. Do not reuse the first probability merely because the colour name is unchanged.

Tree paths without replacement

Tree paths without replacement