A tree tracks trials in order. Branches leaving each junction describe all possible next outcomes, with probabilities adding to 1. Multiply along a complete path; add disjoint paths that match the event. Example A bag contains three red and two blue counters. Draw a counter at random, replace it, mix the bag, then draw again. Write R for red and B for blue. Replacement restores the composition, so each draw has red probability 35 and blue probability 25. 35 -> R -> 25 -> B -> 35 -> R -> 25 -> B -> 35 -> R -> 25 -> B P(RB)= 35 x 25= 625, and P(BR)= 25 x 35= 625. Different colours include both paths, so the probability is 625+ 625= 1225. At least one red is the complement of BB: 1- 25 x 25= 2125. The four terminal probabilities are 925, 625, 625, 425; they sum to 1. A single path does not represent “one of each colour” unless the order is also specified.

A tree tracks trials in order. Branches leaving each junction describe all possible next outcomes, with probabilities adding to \(1\). Multiply along a complete path; add disjoint paths that match the event.

A bag contains three red and two blue counters. Draw a counter at random, replace it, mix the bag, then draw again. Write \(R\) for red and \(B\) for blue. Replacement restores the composition, so each draw has red probability \(\dfrac35\) and blue probability \(\dfrac25\).

\(\mathrm P(RB)=\dfrac35\times\dfrac25=\dfrac6{25}\), and \(\mathrm P(BR)=\dfrac25\times\dfrac35=\dfrac6{25}\).

Different colours include both paths, so the probability is \(\dfrac6{25}+\dfrac6{25}=\dfrac{12}{25}\). At least one red is the complement of \(BB\): \(1-\dfrac25\times\dfrac25=\dfrac{21}{25}\).

The four terminal probabilities are \(\dfrac9{25},\dfrac6{25},\dfrac6{25},\dfrac4{25}\); they sum to \(1\). A single path does not represent “one of each colour” unless the order is also specified.

Tree paths with replacement

Tree paths with replacement