Question

On her vacation, 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?

Answer 1

Veena visits four cities A, B, C and D in a random order.

$\therefore$ Total number of arrangements, Veena can visit four cities A, B, C and D in $4!=24$ ways.

(i) Out of 24 arrangements Veen a can visit city A before B in the following arrangements.

ABCD, ABDC, ACDB, ACBD, ADBC, ADCB, CABD, CADB, CDAB, DABC, DACB, DCAB.

So, there are 12 ways in which Veena can visit city A before city B.

$\therefore$ Required probability = $\frac{12}{24}=\frac{1}{2}.$

(ii) Veena can visit A before B and B before C in any one of the following ways, i.e.

ABCD, ABDC, DABC, ADBC.

So, there are 4 ways in which Veena can visit city A before B and B before C.

$\therefore$ Required probability = $\frac{4}{24}=\frac{1}{6}.$

(iii) Veena can visit city A first and city B last in any one of the following ways, i.e. ACDB, ADCB.

So, there are 2 ways.

$\therefore$ Required probability = $\frac{2}{24}=\frac{1}{2}$