On her vacations,Veena visits four cities (A, | vedclass

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(A) The total number of ways to arrange the four cities $A, B, C$,and $D$ is $4! = 24$. Thus,the total number of outcomes in the sample space is $n(S)

Vedclass · Accepted Answer

$1/6$

Vedclass · Answer

(A) The total number of ways to arrange the four cities $A, B, C$,and $D$ is $4! = 24$. Thus,the total number of outcomes in the sample space is $n(S) = 24$.We are interested in the event $F$ where Veena visits $A$ before $B$ and $B$ before $C$. This means the relative order of $A, B$,and $C$ must be $A 	o B 	o C$.In any permutation of the four cities,there are $3! = 6$ possible relative orders for the cities $A, B$,and $C$. These are:$(A, B, C), (A, C, B), (B, A, C), (B, C, A), (C, A, B), (C, B, A)$.Only one of these orders,$(A, B, C)$,satisfies the condition that $A$ comes before $B$ and $B$ comes before $C$.Since each of these $6$ relative orders is equally likely to occur in the total $24$ permutations,the probability is $\frac{1}{6}$.

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 $C$?

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