Question

On her vacations Veena visits four cities $(A,B,C\text{and}D)$ in a random order. What is the probability that she visits

$\left(i\right)$ $A$ before $B?$

$\left(ii\right)$ $A$ before $B$ and $B$ before $C?$

$\left(iii\right)$ $A$ first and $B$ last $?$

$\left(iv\right)$ $A$ either first or second $?$

$\left(v\right)$ $A$ just before $B?$

Answer 1

$\left(i\right)$ Required Probability
Given : Veena visit four cities $A,B,C,D$ on her vacation in any order.
So, sample space is

$S=\left\{\begin{array}{cccccc}ABCD,& ABDC,& ACBD,& ACDB,& ADBC,& ADCB\\ BACD,& BADC,& BDAC,& BDCA,& BCAD,& BCDA\\ CABD,& CADB,& CBDA,& CBAD,& CDAB,& CDBA\\ DABC,& DACB,& DBCA,& DBAC,& DCAB,& DCBA\end{array}\right\}$

Let $F$ be the event that Veena visit four cities while $A$ before $B$

$\therefore F$

$=\left\{\begin{array}{cccccc}ABCD,& ABDC,& ACBD,& ACDB,& ADBC,& ADCB\\ CABD,& CADB,& CDAB,& DABC,& DACB,& DCAB\end{array}\right\}$

$n\left(S\right)=24\text{and}n\left(F\right)=12$

$P$ $($ Veena visit four cities $A$ before $B$ $)$ $=\frac{\text{favourable outcomes}}{\text{Total outcomes}}$

$P$ $($required event$)$ $=\frac{n\left(F\right)}{n\left(S\right)}=\frac{12}{24}=\frac{1}{2}$

Hence the required probability that Veena visit four cities while $A$ before $B\text{is}\frac{1}{2}$ .

$\left(ii\right)$ Required Probability
Given : Veena visit four cities $A,B,C,D$ on her vacation in any order.
So, sample space is

$S=\left\{\begin{array}{cccccc}ABCD,& ABDC,& ACBD,& ACDB,& ADBC,& ADCB\\ BACD,& BADC,& BDAC,& BDCA,& BCAD,& BCDA\\ CABD,& CADB,& CBDA,& CBAD,& CDAB,& CDBA\\ DABC,& DACB,& DBCA,& DBAC,& DCAB,& DCBA\end{array}\right\}$

Let $F$ be the event that Veena visit four cities while $A$ before $B$ and $B$ before $C$ is

$F=\{ABCD,ABDC,ADBC,DABC\}$

$n\left(S\right)=24\text{and}n\left(F\right)=4$

$P$ $($Veena visit four cities $A$ before $B$ and $B$ before $C$

$=\frac{\text{favourable outcomes}}{\text{Total outcomes}}$

$P$ $($required event$)$ $=\frac{n\left(F\right)}{n\left(S\right)}=\frac{4}{24}=\frac{1}{6}$

Hence the required probability that Veena visit four cities while $A$ before $B$ and $B$ before $C\text{is}\frac{1}{6}$.

$\left(iii\right)$ Required Probability
Given : Veena visit four cities $A,B,C,D$ on her vacation in any order.
So, sample space is

$S=\left\{\begin{array}{cccccc}ABCD,& ABDC,& ACBD,& ACDB,& ADBC,& ADCB\\ BACD,& BADC,& BDAC,& BDCA,& BCAD,& BCDA\\ CABD,& CADB,& CBDA,& CBAD,& CDAB,& CDBA\\ DABC,& DACB,& DBCA,& DBAC,& DCAB,& DCBA\end{array}\right\}$

Let $F$ be the event that Veena visit four cities while $A$ first and $B$ last is

$F=\{ACDB,ADCB\}$

$n\left(S\right)=24\text{and}n\left(F\right)=2$

$P$ $($Veena visit four cities $A$ first and $B$ last$)$

$=\frac{\text{favourable outcomes}}{\text{Total outcomes}}$

$P$ $($required event$)$ $=\frac{n\left(F\right)}{n\left(S\right)}=\frac{2}{24}=\frac{1}{12}$

Hence the required probability that Veena visit four cities $A$ first and $B$ last is $\frac{1}{12}$ .

$\left(iv\right)$ Required Probability
Given : Veena visit four cities on her vacation in any order.
So, sample required sample space is

$S=\left\{\begin{array}{cccccc}ABCD,& ABDC,& ACBD,& ACDB,& ADBC,& ADCB\\ BACD,& BADC,& BDAC,& BDCA,& BCAD,& BCDA\\ CABD,& CADB,& CBDA,& CBAD,& CDAB,& CDBA\\ DABC,& DACB,& DBCA,& DBAC,& DCAB,& DCBA\end{array}\right\}$

Let $F$ be the event that Veena visit four cities while $A$ either first or second is

$F=\left\{\begin{array}{cccc}ABCD,& ABDC,& ACBD,& ACDB\\ ADBC,& ADCB,& BACD,& BADC\\ CABD,& CADB,& DABC,& DACB\end{array}\right\}$

$n\left(S\right)=24\text{and}n\left(F\right)=12$

$P$ $($Veena visit four cities while $A$ either first or second$)$

$=\frac{\text{favourable outcomes}}{\text{Total outcomes}}$

$P$ $($required event$)$ $=\frac{n\left(F\right)}{n\left(S\right)}=\frac{12}{24}=\frac{1}{2}$

Hence the required probability that Veena visit four cities $A$ either first or second is $\frac{1}{2}$.

$\left(v\right)$ Required Probability
Given : Veena visit four cities on her vacation in any order.
So, sample required sample space is

$S=\left\{\begin{array}{cccccc}ABCD,& ABDC,& ACBD,& ACDB,& ADBC,& ADCB\\ BACD,& BADC,& BDAC,& BDCA,& BCAD,& BCDA\\ CABD,& CADB,& CBDA,& CBAD,& CDAB,& CDBA\\ DABC,& DACB,& DBCA,& DBAC,& DCAB,& DCBA\end{array}\right\}$

Let $F$ be the event that Veena visit four cities while $A$ just before $B$ is

$F=\{ABCD,ABDC,CABD,CDAB,DABC,DCAB\}$

$n\left(S\right)=24\text{and}n\left(F\right)=6$

$P$ $($Veena visit four cities while $A$ just before $B)$

$=\frac{\text{favourable outcomes}}{\text{Total outcomes}}$

$P$ $($required event$)$ $=\frac{n\left(F\right)}{n\left(S\right)}=\frac{6}{24}=\frac{1}{4}$

Hence the required probability that Veena visit four cities while $A$ just before $B$ is $\frac{1}{4}$.