On her vacations Veena visits four cities $\left(A,B,CandD\right)$in a random order. What is the probability that she visits(i) A before B?(ii) A before B and B before C?(iii) A first and B last?(iv) A either first or second?(v) A just be

Solution: Here, sample space will contain $4!$ ways.
$S=$
$\{ABCD,ACBD,ABDC,ADBC,ACDB,ADCB,$
$BACD,BADC,BCAD,BCDA,BDAC,BDCA,$
$CABD,CADB,CBAD,CBDA,CDAB,CDBA,$
$DABC,DACB,DBAC,DBCA,DCAB,DCBA\}$
So, $n\left(s\right)=24$
(i)Number of ways Veena travels A before B $=12$
$\therefore$ Probability of her traveling A before B$=\frac{12}{24}=\frac{1}{2}$

(ii) Number of ways Veena travels A before B and B before C$=4$
$\therefore$ Probability of her traveling A before B and B before C$=\frac{4}{24}=\frac{1}{6}$

(iii)Number of ways Veena travels A first and B last$=2$
$\therefore$ Probability of her traveling A first and B last $=\frac{2}{24}=\frac{1}{12}$

(iv)Number of ways Veena travels A either first or second $=12$
$\therefore$ Probability of her traveling A either first or second$=\frac{12}{24}=\frac{1}{2}$

(v)Number of ways Veena travels A just before B $=6$
$\therefore$ Probability of her traveling A just before B $=\frac{6}{24}=\frac{1}{4}$