On her vacations Veena visits four cities (A, B, C and D) in a random order. What is the probability that she visits (i) A before B? (iii) A first and B last? (v) A just before B? (ii) $\mathrm{A}$ before $\mathrm{B}$ and $\mathrm{B}$ before $\mathrm{C}$ ? (iv) A either first or second?

Step 1: Calculate Total Possible Arrangements

Veena visits four cities (A, B, C, D) in a random order. The total number of ways to arrange these four distinct cities is given by the factorial of the number of cities, which is $4!$.

Step 2: Calculate Probability for (i) A before B

For any two specific cities, say A and B, in any random arrangement of the four cities, A will appear before B in exactly half of the arrangements, and B will appear before A in the other half. Thus, the probability is $1/2$.

Step 3: Calculate Probability for (ii) A before B and B before C

For any three specific cities (A, B, C), there are $3! = 6$ possible orders in which they can appear. Only one of these orders is A before B before C. So, the probability is $1/3! = 1/6$.

Step 4: Calculate Probability for (iii) A first and B last

If A is first and B is last, the remaining two cities (C and D) can be arranged in the two middle positions in $2!$ ways. The total number of arrangements is $4!$. So the probability is $2!/4!$.

Step 5: Calculate Probability for (iv) A either first or second

If A is first, the remaining 3 cities can be arranged in $3!$ ways. If A is second, the remaining 3 cities can be arranged in $3!$ ways. Since these are mutually exclusive events, we add their probabilities. Alternatively, A can be in any of the 4 positions with equal probability, so the probability of A being in the first or second position is $2/4 = 1/2$.

Step 6: Calculate Probability for (v) A just before B

Consider 'AB' as a single block. Now we are arranging 3 items: (AB), C, and D. These 3 items can be arranged in $3!$ ways. The total number of arrangements is $4!$. So the probability is $3!/4!$.

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