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

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(B) The total number of ways to visit $4$ cities is $4! = 4 	imes 3 	imes 2 	imes 1 = 24$. Let $S$ be the sample space of all possible permutations

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(B) The total number of ways to visit $4$ cities is $4! = 4 	imes 3 	imes 2 	imes 1 = 24$. Let $S$ be the sample space of all possible permutations of ${A, B, C, D}$. $n(S) = 24$. Let $H$ be the event that she visits $A$ either first or second. If $A$ is visited first,the remaining $3$ cities can be visited in $3! = 6$ ways. If $A$ is visited second,the remaining $3$ cities can be visited in $3! = 6$ ways. Thus,the number of favorable outcomes is $n(H) = 6 + 6 = 12$. The probability $P(H)$ is given by: $P(H) = \frac{n(H)}{n(S)} = \frac{12}{24} = \frac{1}{2}$.

On her vacations,Veena visits four cities ($A, B, C$,and $D$) in a random order. What is the probability that she visits $A$ either first or second?

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