On her vacations Veena visits four cities $(A,\,B ,\, C$ and $D$ ) in a random order. What is the probability that she visits $A$ before $B$ and $B$ before $C ?$
On her vacations Veena visits four cities $(A,\,B ,\, C$ and $D$ ) in a random order. What is the probability that she visits $A$ before $B$ and $B$ before $C ?$
The number of arrangements (orders) in which Veena can visit four cities $A,\,B,\,C$ or $D$ is $4 !$ i.e., $24 .$ Therefore, $n(S)=24$
since the number of elements in the sample space of the experiment is $24$ all of these outcomes are considered to be equally likely. A sample space for the experiment is
$S =\{ ABCD , \,ABDC , \,ACBD $, $ACDB , \,ADBC , \,ADCB$, $BACD,\, BADC,\, BDAC$, $BDCA, \,BCAD, ,BCDA,$ $CABD, \,CADB, \,CBDA$, $CBAD, \,CDAB, \,CDBA,$ $DABC,\, DACB,\, DBCA$, $DBAC, \,DCAB, \,DCBA\}$
Let the event 'Veena visits A before $B$ and $B$ before $C ^{*}$ be denoted by $F$.
Here $F =\{ ABCD , \,DABC , \,ABDC , \,ADBC \}$
Therefore, $P(F)=\frac{n(F)}{n(S)}=\frac{4}{24}=\frac{1}{6}$
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