VT · Probability Trees

Non-calculator

In each set, one thing changes and everything else stays the same. Work them out in order and look for the pattern — the last line tells you what to notice.

①Independent events, two goes – the probability of winning changeschanging: P(win)

For each probability of winning a single go, work out P(win on both goes).

P(win) $=\dfrac{1}{2}$

P(win) $=\dfrac{1}{3}$

P(win) $=\dfrac{2}{5}$

P(win) $=\dfrac{3}{4}$

P(win and win) $= $ P(win) $\times$ P(win). Why do you multiply the two probabilities?

②Without replacement – a bag of 5 counters, the number of red counters changeschanging: number of red counters (out of 5)

A bag holds 5 counters. Two are taken without replacement. For each number of red counters, work out P(both red).

2 red

3 red

4 red

5 red

The second fraction has a smaller denominator because you do not replace the first counter. How do both the top and bottom change?

③Same experiment (2 red, 3 blue, take 2 without replacement) – the question changeschanging: what you are asked to find

A bag holds 2 red and 3 blue counters. Two are taken without replacement. Work out each probability.

P(both red)

P(both blue)

P(one of each colour)

P(at least one red)

Same tree, different questions. Which of these are quicker using $1-$ P(something)?

Answers · Probability Trees

Variation practice

① Independent events, two goes – the probability of winning changes

P(win) $=\dfrac{1}{2}$: $\dfrac{1}{4}$P(win) $=\dfrac{1}{3}$: $\dfrac{1}{9}$P(win) $=\dfrac{2}{5}$: $\dfrac{4}{25}$P(win) $=\dfrac{3}{4}$: $\dfrac{9}{16}$

② Without replacement – a bag of 5 counters, the number of red counters changes

2 red: $\dfrac{1}{10}$3 red: $\dfrac{3}{10}$4 red: $\dfrac{3}{5}$5 red: 1

③ Same experiment (2 red, 3 blue, take 2 without replacement) – the question changes

P(both red): $\dfrac{1}{10}$P(both blue): $\dfrac{3}{10}$P(one of each colour): $\dfrac{3}{5}$P(at least one red): $\dfrac{7}{10}$