Answer
a) $\frac{2}{3}$
b) $\frac{3}{4}$
Work Step by Step
Let E=Late
$F_1$=Car
$F_2$=Bus
$F_3$=Bicycle
$P(E|F_1)=50\%=0.5$
$P(E|F_2)=20\%=0.2$
$P(E|F_3)=5\%=0.05$
a)
$P(F_1)=\frac{1}{3}$
$P(F_2)=\frac{1}{3}$
$P(F_3)=\frac{1}{3}$
Using the generalized bayes' theorem:
$P(F_1|E)=\displaystyle \frac{P(E|F_1)P(F_1)}{P(E|F_1)P(F_1)+P(E|F_2)P(F_2)+P(E|F_3)P(F_3)}$
$\displaystyle=\frac{0.5.(1/3)}{0.5\times (1/3)+0.2\times (1/3)+0.05\times(1/3)}=\frac{2}{3} \approx 0.6667$
b)
$P(F_1)=30\%=0.3$
$P(F_2)=10\%=0.1$
$P(F_3)=60\%=0.6$
Using the generalized bayes' theorem:
$P(F_1|E)=\displaystyle \frac{P(E|F_1)P(F_1)}{P(E|F_1)P(F_1)+P(E|F_2)P(F_2)+P(E|F_3)P(F_3)}$
$\displaystyle=\frac{0.5.\times 0.3}{0.5\times 0.3+0.2\times 0.1+0.05\times 0.6}=\frac{3}{4} =0.75$
