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) A before B and B before C ?
(iv) A either first or second?
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Step-by-Step Solution
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!.