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 ii) A before B and B before C iii)A first and B last iv)A either first or second v)A just before B.

The number of arrangements (orders) in which Veena can visit four cities A,
B, C, or D is 4! i.e., 24.Therefore, n (S) = 24.
Since the number of elements in the sample space of the experiment is 24 all of these
outcomes are considered to be equally likely. A 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}


(iii) A first and B last
Here there are two cases : (ACDB) and (ADCB)
So the probability of this event is 2/24 = 1/12

(iv) A either first or second
Here there are 12 cases of A either first or second.
So the probability of the event is 12/24 = 1/2

Try the last part yourself.

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