PLZ EXPLAINN EX NO-14OF PROBABILITY

Here is the solution.

The number of arrangements (orders) in which Veena can visit four cities A, B, C, or D is 4! = 24.

Therefore, n (S) = 24.          (S = sample space)
Since the number of elements in the sample space of the experiment is 24 and all of these outcomes are considered to be equally likely. 

The sample space for the experiment is:
S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC, BDAC, BDCA, BCAD, BCDA, CABD, CADB, CBDA, CBAD, CDAB, CDBA, DABC, DACB, DBCA, DBAC, DCAB, DCBA}
 

(i) Let the event ‘she visits A before B’ be denoted by E
Therefore, favourable cases for E = {ABCD, CABD, DABC, ABDC, CADB, DACB, ACBD, ACDB, ADBC, CDAB, DCAB, ADCB}
Thus  
 

(ii) Let the event ‘Veena visits A before B and B before C’ be denoted by F.
The favourable cases for F = {ABCD, DABC, ABDC, ADBC}
Therefore,

In the similar manner, we can solve other parts.

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