- Home
- Standard 11
- Mathematics
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 $B$ ?
$\frac{1}{2}$
$\frac{1}{2}$
$\frac{1}{2}$
$\frac{1}{2}$
Solution
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 'she visits $A$ before $B ^{\prime}$ be denoted by $E$
Therefore,
$E =\{ ABCD ,\, CABD$ ,$ DABC ,\, ABDC$ , $CADB ,\, DACB$ $ACBD ,\, ACDB , ADBC $, $CDAB ,\, DCAB ,\, ADCB \}$
Thus $P ( E )=\frac{n( E )}{n( S )}=\frac{12}{24}=\frac{1}{2}$
