Conditional probability and independence of events

a) The condition is "Sales": the universe is the 30 people in that row, and 18 work from home. $$P(\text{works from home}\mid \text{Sales})=\frac{18}{30}=\frac{3}{5}$$

b) The condition is "works from home": the column of 38 people, of whom 18 are in Sales. $$P(\text{Sales}\mid \text{works from home})=\frac{18}{38}=\frac{9}{19}$$

c) Condition "does not work from home": 42 people, of whom 30 are in the Technical department. $$P(\text{Technical}\mid \text{does not work from home})=\frac{30}{42}=\frac{5}{7}$$

d) We compare $P(V\cap T)$ and $P(V)\cdot P(T)$, with $V$ "Sales" and $T$ "works from home": $$P(V\cap T)=\frac{18}{80}=\frac{9}{40}=\frac{72}{320}P(V)\cdot P(T)=\frac{30}{80}\cdot \frac{38}{80}=\frac{57}{320}$$ They do not coincide, so they are not independent. Consistent with a), $P(T\mid V)=\frac{3}{5}=0.6$, while $P(T)=\frac{38}{80}=0.475$.

Answer: a) $\frac{3}{5}$; b) $\frac{9}{19}$; c) $\frac{5}{7}$; d) not independent.

a) Product rule: $$P(A\cap B)=P(A\mid B)\cdot P(B)=\frac{3}{4}\cdot \frac{2}{5}=\frac{3}{10}$$

b) The condition is now $A$: $$P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{3/10}{7/10}=\frac{3}{7}$$

c) Union formula: $$P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac{7}{10}+\frac{4}{10}-\frac{3}{10}=\frac{4}{5}$$

d) The cases of $A$ that are not in $B$ are $A\cap B^{\prime }$, with $P(A\cap B^{\prime })=P(A)-P(A\cap B)=\frac{7}{10}-\frac{3}{10}=\frac{2}{5}$, and $P(B^{\prime })=1-\frac{2}{5}=\frac{3}{5}$. Then $$P(A\mid B^{\prime })=\frac{2/5}{3/5}=\frac{2}{3}$$

e) Compare: $P(A)\cdot P(B)=\frac{7}{10}\cdot \frac{2}{5}=\frac{7}{25}$ and $P(A\cap B)=\frac{3}{10}$. As $\frac{7}{25}\ne \frac{3}{10}$, they are not independent. This is also seen in the fact that $P(A\mid B)=\frac{3}{4}\ne \frac{7}{10}=P(A)$.

Answer: a) $\frac{3}{10}$; b) $\frac{3}{7}$; c) $\frac{4}{5}$; d) $\frac{2}{3}$; e) no.

Let $a=P(A)$ and $b=P(B)$. Since $A$ and $B$ are independent, so are $A$ with $B^{\prime }$ and $A^{\prime }$ with $B$, so $$a(1-b)=\frac{1}{3}(1-a)b=\frac{1}{6}$$

a) Subtracting the two equations, the $ab$ terms cancel: $a-b=\frac{1}{3}-\frac{1}{6}=\frac{1}{6}$, that is, $a=b+\frac{1}{6}$. Substituting into the first: $$\left(b+\frac{1}{6}\right)(1-b)=\frac{1}{3}\Rightarrow -b^2+\frac{5}{6}b+\frac{1}{6}=\frac{1}{3}\Rightarrow 6b^2-5b+1=0$$ Factorising, $(2b-1)(3b-1)=0$, so $b=\frac{1}{2}$ or $b=\frac{1}{3}$.

b) If $b=\frac{1}{2}$, then $a=\frac{2}{3}$; if $b=\frac{1}{3}$, then $a=\frac{1}{2}$. Both pairs satisfy the two equations. The condition $P(A)<\frac{3}{5}$ rules out $a=\frac{2}{3}$ (since $\frac{2}{3}>\frac{3}{5}$), so $$P(A)=\frac{1}{2}P(B)=\frac{1}{3}$$

c) With independence, $P(A\cap B)=ab=\frac{1}{2}\cdot \frac{1}{3}=\frac{1}{6}$, and $$P(A\cup B)=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{2}{3}$$

d) From the calculation above, $P(A\cap B)=\frac{1}{6}$. Check: $P(A\cap B^{\prime })=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}$ and $P(A^{\prime }\cap B)=\frac{1}{3}-\frac{1}{6}=\frac{1}{6}$, as the data says.

In the exam: marks are awarded for writing that independence allows you to use $P(A\cap B^{\prime })=P(A)P(B^{\prime })$ and for setting up the two equations. If, when solving the quadratic, you keep only one root without explaining why, you lose the justification mark: here both roots are valid and the extra piece of data is needed to decide.