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.

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.

**
**