Statistics

Tree diagrams

Follow two draws, with or without replacement.

Predict, test and explain

Step 1 of 3

Predict

A bag has 3 red and 2 blue counters. Without replacement, what is P(red, red)?

P(RR)=\frac{3}{5}\times\frac{1}{2}=\frac{3}{10}

Scroll the diagram sideways to see every label.

\sum P(\text{paths})=1

P(RB\text{ or }BR)=\frac{3}{5}

Two draws without replacement. The second draw uses one fewer counter. Square counters show the starting bag.

All two-draw paths
Path	First draw	Second draw	Probability
Red, Red	$\frac{3}{5}$	$\frac{1}{2}$	$\frac{3}{10}$
Red, Blue	$\frac{3}{5}$	$\frac{1}{2}$	$\frac{3}{10}$
Blue, Red	$\frac{2}{5}$	$\frac{3}{4}$	$\frac{3}{10}$
Blue, Blue	$\frac{2}{5}$	$\frac{1}{4}$	$\frac{1}{10}$

Explore and compare

Choose the investigation above. Predict what will change before you move a control.

Predict

Try it in the simulation. Change one thing at a time.

Explain

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Hints, self-check and connections

Try a self-check

Fixed values, separate from the diagram controls.

A bag has 4 red and 2 blue counters. Two are drawn without replacement. Find the probability of two reds.

$P(RR)=?$

Hint

After the first red is kept out, there are 3 reds among 5 counters.

Reveal answer

Multiply along the red-red path. The probability is two fifths.

$P(RR)=\frac46\times\frac35=\frac25$

Hint 1

Read whether the first counter is replaced before the second draw.

Hint 2

Multiply probabilities along one complete path.

Hint 3

Without replacement, update the remaining colour counts and total; add separate paths for an either/or event.

Worked example

This example uses fixed values, separate from the diagram controls.

A bag has 3 red and 2 blue counters. Find P(red then red) without replacement.

Answer: 3/10.

Connect this idea

Watch out: Without replacement, the second probability changes; the second denominator is one smaller.