Total probability, Bayes' theorem and tree diagrams

Let $F$ be "strawberry" and $L$ be "lemon". Without returning, 7 sweets are left at the second stage.

a) Multiply along the path $FF$: $$P(FF)=\frac{3}{8}\cdot \frac{2}{7}=\frac{6}{56}=\frac{3}{28}$$

b) "Same flavour" means the paths $FF$ and $LL$, which are added: $$P(LL)=\frac{5}{8}\cdot \frac{4}{7}=\frac{20}{56}=\frac{5}{14}P(\text{same flavour})=\frac{3}{28}+\frac{10}{28}=\frac{13}{28}$$

c) With replacement the probabilities repeat: $$P(FF)=\frac{3}{8}\cdot \frac{3}{8}=\frac{9}{64}$$

d) Two paths end in "second is strawberry": $FF$ and $LF$. $$P=\frac{3}{8}\cdot \frac{2}{7}+\frac{5}{8}\cdot \frac{3}{7}=\frac{6}{56}+\frac{15}{56}=\frac{21}{56}=\frac{3}{8}$$ This is the same as the probability for the first sweet: with no information about the first draw, there is no reason for the second to be different.

Answer: a) $\frac{3}{28}$; b) $\frac{13}{28}$; c) $\frac{9}{64}$; d) $\frac{3}{8}$.

Let $B$ be "bus", $P$ be "walk", $C$ be "bicycle" and $T$ be "arrives late". The data are $P(B)=\frac{2}{5}$, $P(P)=\frac{7}{20}$ and $P(C)=1-\frac{2}{5}-\frac{7}{20}=\frac{1}{4}$. At the second stage: $P(T\mid B)=\frac{1}{10}$, $P(T\mid P)=\frac{1}{25}$ and $P(T\mid C)=\frac{1}{50}$.

a) Three paths end in $T$; they are added (total probability): $$P(T)=\frac{2}{5}\cdot \frac{1}{10}+\frac{7}{20}\cdot \frac{1}{25}+\frac{1}{4}\cdot \frac{1}{50}=\frac{40}{1000}+\frac{14}{1000}+\frac{5}{1000}=\frac{59}{1000}$$

b) Among the cases that end in $T$ (denominator $\frac{59}{1000}$), the bus path is worth $\frac{40}{1000}$: $$P(B\mid T)=\frac{P(B\cap T)}{P(T)}=\frac{40/1000}{59/1000}=\frac{40}{59}$$

c) Now the possible cases are those that end in $T^{\prime }$, with $P(T^{\prime })=1-\frac{59}{1000}=\frac{941}{1000}$. The path "cycle and not late" is worth $$P(C\cap T^{\prime })=\frac{1}{4}\cdot \frac{49}{50}=\frac{49}{200}=\frac{245}{1000}$$ and then $$P(C\mid T^{\prime })=\frac{245/1000}{941/1000}=\frac{245}{941}$$

Answer: a) $\frac{59}{1000}$; b) $\frac{40}{59}$; c) $\frac{245}{941}$.

Let $E$ be "ill" and $+$ "positive test". Data: $P(E)=\frac{1}{50}$, $P(+\mid E)=\frac{19}{20}$ and $P(+\mid E^{\prime })=\frac{1}{10}$. The missing branches: $P(E^{\prime })=\frac{49}{50}$, $P(-\mid E)=\frac{1}{20}$ and $P(-\mid E^{\prime })=\frac{9}{10}$.

a) Two paths end in a positive result: $$P(+)=\frac{1}{50}\cdot \frac{19}{20}+\frac{49}{50}\cdot \frac{1}{10}=\frac{19}{1000}+\frac{98}{1000}=\frac{117}{1000}$$

b) Bayes' theorem (or ratio of paths): $$P(E\mid +)=\frac{19/1000}{117/1000}=\frac{19}{117}\approx 0.162$$ Although the test detects 95% of ill people, only 16.2% of positives are ill: the disease is so rare that the false positives from the large healthy majority weigh more.

c) With independence within each state, the probability of two positives in a row for an ill person is ${\left(\frac{19}{20}\right)}^2=\frac{361}{400}$, and for a healthy one ${\left(\frac{1}{10}\right)}^2=\frac{1}{100}$. The paths that end in "two positives" are worth $$P(E\cap ++)=\frac{1}{50}\cdot \frac{361}{400}=\frac{361}{20000}P(E^{\prime }\cap ++)=\frac{49}{50}\cdot \frac{1}{100}=\frac{196}{20000}$$ and then $$P(E\mid ++)=\frac{361/20000}{(361+196)/20000}=\frac{361}{557}\approx 0.648$$ A second positive result raises the probability from 16.2% to 64.8%.

d) The possible cases are those that end in a negative result: $P(-)=1-\frac{117}{1000}=\frac{883}{1000}$. The path "healthy and negative" is worth $\frac{49}{50}\cdot \frac{9}{10}=\frac{441}{500}=\frac{882}{1000}$. Therefore $$P(E^{\prime }\mid -)=\frac{882/1000}{883/1000}=\frac{882}{883}\approx 0.999$$

In the exam: marks are given for leaving the tree (or the products of the branches) and for showing the quotient with numerator and denominator separately. The classic mistake is to answer b) with $\frac{19}{20}$, which is $P(+\mid E)$, the probability the wrong way round. In c) you must justify why the probabilities of the two tests are multiplied within each branch; without that step you cannot continue.