Question

On her vacation veena visit four cities A B C D what is the probability that she visit A first and B last

Answer 1

total no of cities is 4

the probability of visiting A first is 1/4

the probability of visiting B last is 1/4

I hope it will help you

Answer 2

Total cities are 4 i.e. A, B, C, D

Given, Veena visit four cities, So, n(S) = 4! = 24

Now, sample space IS:

S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB,

BACD, BADC, BDAC, BDCA, BCAD, BCDA,

CABD, CADB, CBDA, CDAD, CDAB,CDBA,

DABC, DACB, DBCA, DBAC, DCAB, DCBA}

1. Let E = Veena visits A before B

=> E = {ABCD, CABD, DABC, ABDC, CADB, DACB, ACBD, ACDB, ADBC, CDAB, DCAB, ADCB}

=> n(E) = 12

So, P(E) = n(E)/n(S) = 12/24 = 1/2

2. Let F = Veena visits A before B and B before C

=> F = {ABCD, DABC, ABDC, ADBC}

=> n(F) = 4

So, P(F) = n(F)/n(S) = 4/24 = 1/6

3. Let G = Veena visits A firsta and B last

=> G = {ACDB, ADCB}

=> n(G) = 2

So, P(G) = n(G)/n(S) = 2/24 = 1/12

4. Let H = Veena visits A either first or second

=> Total number of cases = 12

=> n(H) = 12

So, P(H) = n(H)/n(S) = 12/24 = 1/2

5. Let I = Veena visits A just before B

=> I = {ABCD, ABDC, CABD, DABC, CDAB, DCAB}

=> n(I) = 6

So, P(I) = n(I)/n(S) = 6/24 = 1/4