Challenge · Probability Trees

No calculator20 marks

These problems combine ideas – you will plan your own steps, and for the harder ones form your own equation. There is a worked example to show you how to start; then it is over to you. Stuck? Hints are at the foot of the sheet. No calculator.

Worked example: how to start
A spinner lands on red with probability $0.6$. It is spun twice. Work out the probability that it lands on red exactly once.
Exactly once means red-then-not-red OR not-red-then-red: add the two paths.
P(red, not red) $=0.6\times0.4=0.24$ and P(not red, red) $=0.4\times0.6=0.24$.
Total $=0.24+0.24=0.48$.
1A bag contains $5$ green and $3$ yellow sweets. Two sweets are taken at random without replacement. Work out the probability that the two sweets are the same colour.[4 marks]
Answer:
2In a game you play two rounds. The probability of winning the first round is $\dfrac{2}{3}$. If you win the first round, the probability of winning the second is $\dfrac{1}{2}$; if you lose the first, the probability of winning the second is $\dfrac{1}{4}$. Work out the probability that you win exactly one of the two rounds.[4 marks]
Answer:
3It rains on any given day with probability $\dfrac{1}{5}$, independently of other days. Work out the probability that it rains on at least one of the next two days.[3 marks]
Answer:
4A biased coin lands on heads with probability $p$. It is flipped twice. The probability of getting two heads is $\dfrac{9}{25}$. Work out the probability of getting two tails.[4 marks]
Answer:
5A bag contains $n$ red counters and $4$ blue counters. Two counters are taken at random without replacement. The probability that both counters are blue is $\dfrac{1}{6}$. Work out the value of $n$.[5 marks]
Answer:
★Extension: A bag holds $3$ red and $2$ blue counters. Counters are taken one at a time, without replacement, until a blue counter is taken. Show that the probability that this takes exactly two draws is $\dfrac{3}{10}$.
Stuck? Hints (don't peek unless you need to)1. Same colour means both green OR both yellow. Work out each probability along its branches, then add them.2. The second-round probability depends on the first – this is a conditional tree. Exactly one win means win-then-lose OR lose-then-win.3. Use $1-$ P(no rain on either day). No rain has probability $\dfrac{4}{5}$ each day.4. First find $p$ from $p^2=\dfrac{9}{25}$, then the probability of tails is $1-p$.5. Write P(both blue) $=\dfrac{4}{n+4}\times\dfrac{3}{n+3}$, set it equal to $\dfrac{1}{6}$, and solve the resulting quadratic. Reject any negative value.

Solutions & mark scheme · Probability Trees

Total: 20 marks

Award the marks shown for each correct step; many of these have more than one valid route, so give method marks for any correct working.

1A bag contains $5$ green and $3$ yellow sweets. Two sweets are taken at random without replacement. Work out the probability that the two sweets are the same colour.[4]
Model solution
P(green, green) $=\dfrac{5}{8}\times\dfrac{4}{7}=\dfrac{20}{56}=\dfrac{5}{14}$.
P(yellow, yellow) $=\dfrac{3}{8}\times\dfrac{2}{7}=\dfrac{6}{56}=\dfrac{3}{28}$.
Same colour $=\dfrac{5}{14}+\dfrac{3}{28}=\dfrac{10}{28}+\dfrac{3}{28}=\dfrac{13}{28}$.
Answer: $\dfrac{13}{28}$
Marks
✔1P(GG) $=\frac58\times\frac47=\frac{5}{14}$
✔1P(YY) $=\frac38\times\frac27=\frac{3}{28}$
✔1Adds the two same-colour probabilities
✔1$\frac{13}{28}$
2In a game you play two rounds. The probability of winning the first round is $\dfrac{2}{3}$. If you win the first round, the probability of winning the second is $\dfrac{1}{2}$; if you lose the first, the probability of winning the second is $\dfrac{1}{4}$. Work out the probability that you win exactly one of the two rounds.[4]
Model solution
P(win, lose) $=\dfrac{2}{3}\times\dfrac{1}{2}=\dfrac{1}{3}$.
P(lose, win) $=\dfrac{1}{3}\times\dfrac{1}{4}=\dfrac{1}{12}$.
Exactly one win $=\dfrac{1}{3}+\dfrac{1}{12}=\dfrac{4}{12}+\dfrac{1}{12}=\dfrac{5}{12}$.
Answer: $\dfrac{5}{12}$
Marks
✔1P(win, lose) $=\frac23\times\frac12=\frac13$
✔1P(lose, win) $=\frac13\times\frac14=\frac{1}{12}$
✔1Adds the two paths
✔1$\frac{5}{12}$
3It rains on any given day with probability $\dfrac{1}{5}$, independently of other days. Work out the probability that it rains on at least one of the next two days.[3]
Model solution
P(no rain on a day) $=1-\dfrac{1}{5}=\dfrac{4}{5}$.
P(no rain on both days) $=\dfrac{4}{5}\times\dfrac{4}{5}=\dfrac{16}{25}$.
P(at least one rainy day) $=1-\dfrac{16}{25}=\dfrac{9}{25}$.
Answer: $\dfrac{9}{25}$
Marks
✔1P(no rain both) $=\left(\frac45\right)^2=\frac{16}{25}$
✔1Uses $1-\frac{16}{25}$
✔1$\frac{9}{25}$
4A biased coin lands on heads with probability $p$. It is flipped twice. The probability of getting two heads is $\dfrac{9}{25}$. Work out the probability of getting two tails.[4]
Model solution
$p^2=\dfrac{9}{25}$, so $p=\dfrac{3}{5}$.
P(tails) $=1-\dfrac{3}{5}=\dfrac{2}{5}$.
P(two tails) $=\left(\dfrac{2}{5}\right)^2=\dfrac{4}{25}$.
Answer: $\dfrac{4}{25}$
Marks
✔1$p^2=\frac{9}{25}$
✔1$p=\frac35$
✔1P(tails) $=\frac25$
✔1$\frac{4}{25}$
5A bag contains $n$ red counters and $4$ blue counters. Two counters are taken at random without replacement. The probability that both counters are blue is $\dfrac{1}{6}$. Work out the value of $n$.[5]
Model solution
P(both blue) $=\dfrac{4}{n+4}\times\dfrac{3}{n+3}=\dfrac{12}{(n+4)(n+3)}$.
Set equal to $\dfrac{1}{6}$: $\dfrac{12}{(n+4)(n+3)}=\dfrac{1}{6}$, so $(n+4)(n+3)=72$.
$n^2+7n+12=72$, so $n^2+7n-60=0$, i.e. $(n+12)(n-5)=0$.
$n=5$ (reject $n=-12$).
Answer: n = 5
Marks
✔1P(BB) $=\frac{4}{n+4}\times\frac{3}{n+3}$
✔1Equation $\frac{12}{(n+4)(n+3)}=\frac16$
✔1$(n+4)(n+3)=72$
✔1$n^2+7n-60=0$
✔1$n=5$
★Extension
Exactly two draws means the first counter is red and the second is blue.
P(red first) $=\dfrac{3}{5}$; then $4$ counters remain, $2$ of them blue, so P(blue second) $=\dfrac{2}{4}$.
P(exactly two draws) $=\dfrac{3}{5}\times\dfrac{2}{4}=\dfrac{6}{20}=\dfrac{3}{10}$.
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