A tourist visits four cities A, B, C and D in a random order. What is the proba…

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A tourist visits four cities A, B, C and D in a random order. What is the probability that he visits A before B?

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Out of all $4! = 24$ permutations, in half of them A comes before B, and in the other half B comes before A. Therefore probability = $\dfrac{1}{2}$.

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