Represent compound events using tree diagrams

Tree diagrams represent the outcomes of a compound experiment as branches, with each complete path showing one possible sequence and the associated probabilities at each stage. The probability of a path is found by multiplying successive branch probabilities, and the probability of an event is found by adding the probabilities of the relevant, nonoverlapping paths; branch probabilities may change when outcomes are dependent. The scope is limited to straightforward two-stage or small multi-stage experiments, not advanced conditional-probability theory or generalized stochastic models.

Detailed Explanation: Represent compound events using tree diagrams

A tree diagram shows the possible outcomes of a compound experiment in stages.

Each branch represents one possible outcome.
Write the probability on each branch.
Multiply probabilities along one complete path.
Add the probabilities of paths that make up the event you want.

Example

A bag contains 3 red counters and 2 blue counters. One counter is drawn, not replaced, and then a second counter is drawn. What is the probability of drawing one red and one blue?

Step 1: Draw the first branches

There are $5$ counters altogether:

P(R)=\frac{3}{5}, \qquad P(B)=\frac{2}{5}

The first branches are:

Start
├── Red     3/5
└── Blue    2/5

Step 2: Add the second-stage branches

Because the first counter is not replaced, the probabilities change.

If the first counter is red, there are $2$ red and $2$ blue counters left.
If the first counter is blue, there are $3$ red and $1$ blue counter left.

Start
├── Red 3/5
│   ├── Red  2/4
│   └── Blue 2/4
└── Blue 2/5
    ├── Red  3/4
    └── Blue 1/4

Each complete path shows one possible sequence:

Red, Red
Red, Blue
Blue, Red
Blue, Blue

Step 3: Find the probability of each successful path

“One red and one blue” can happen in two ways:

Red, then Blue

P(RB)=\frac{3}{5}\times\frac{2}{4} =\frac{6}{20} =\frac{3}{10}

Blue, then Red

P(BR)=\frac{2}{5}\times\frac{3}{4} =\frac{6}{20} =\frac{3}{10}

Step 4: Add the successful paths

The paths $RB$ and $BR$ cannot happen at the same time, so add their probabilities:

P(\text{one red and one blue}) =\frac{3}{10}+\frac{3}{10} =\frac{6}{10} =\boxed{\frac{3}{5}}

So, the probability of drawing one red and one blue is $\boxed{\frac{3}{5}}$ .

Learn by doing: Represent compound events using tree diagrams

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