Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Last updated at May 29, 2023 by Teachoo
Example 9 On her vacations Veena visits four cities (A, B, C and D) in a random order. What is the probability that she visits A before B? 4 cities can be visited in any of following order S = {β("ABCD, ABDC, ACBD, ACDB, ADBC, ADCB," @" BACD, BADC, BDAC, BDCA, BCAD, BCDA," @" CABD, CADB, CBDA, CBAD, CDAB, CDBA," @" DABC, DACB, DBCA, DBAC, DCAB, DCBA" )} n(S) = 24 Let E be the event that βshe visits A before Bβ Hence , E = {β("ABCD, ABDC, ADBC, ACDB, ADBC, ADCB," @"CABD, CADB, CDAB, " @"DABC, DACB, DCAB," )} n(E) = 12 P(E) = (π(πΈ))/(π(π)) = 12/24 = π/π Example 14 What is the probability that she visits (ii) A before B and B before C? S = {β("ABCD, ABDC, ACBD, ACDB, ADBC, ADCB," @" BACD, BADC, BDAC, BDCA, BCAD, BCDA," @" CABD, CADB, CBDA, CBAD, CDAB, CDBA," @" DABC, DACB, DBCA, DBAC, DCAB, DCBA" )} Let F be the event βshe visits A before B and B before C β F = {β("ABCD, ABDC, ADBC , DABC" )} So, n(F) = 4 P(F) = (π(πΉ))/(π(π)) = 4/24 = π/π Example 14 What is the probability that she visits (iii) A first and B last? S = {β("ABCD, ABDC, ACBD, ACDB, ADBC, ADCB," @" BACD, BADC, BDAC, BDCA, BCAD, BCDA," @" CABD, CADB, CBDA, CBAD, CDAB, CDBA," @" DABC, DACB, DBCA, DBAC, DCAB, DCBA" )} Let G be the event βshe visits A first and B lastβ G = {β("ACDB, ADCB" )} So, n(G) = 2 P(G) = (π(πΊ))/(π(π)) = 2/24 = π/ππ Example 14 What is the probability that she visits (iv) A either first or second? S = {β("ABCD, ABDC, ACBD, ACDB, ADBC, ADCB," @" BACD, BADC, BDAC, BDCA, BCAD, BCDA," @" CABD, CADB, CBDA, CBAD, CDAB, CDBA," @" DABC, DACB, DBCA, DBAC, DCAB, DCBA" )} Let H be the event βshe visits A either first or secondβ H = {β("ABCD, ABDC, ADBC, ACDB, ADBC, ADCB," @" BACD, BADC,CABD, CADB,DABC, DACB," )} So, n(H) = 12 P(H) = (π(π»))/(π(π)) = 12/24 = π/π Example 14 What is the probability that she visits (v) A just before B? S = {β("ABCD, ABDC, ACBD, ACDB, ADBC, ADCB," @" BACD, BADC, BDAC, BDCA, BCAD, BCDA," @" CABD, CADB, CBDA, CBAD, CDAB, CDBA," @" DABC, DACB, DBCA, DBAC, DCAB, DCBA" )} Let I be the event βshe visits A just before Bβ I = {β("ABCD, ABDC, CABD, CDAB, DABC, DCAB," )} So, n(I) = 6 P(I) = (π(πΌ))/(π(π)) = 6/24 = π/π