Example 14 - Chapter 16 Class 11 Probability (Term 2)
Last updated at Feb. 11, 2020 by Teachoo
Last updated at Feb. 11, 2020 by Teachoo
Transcript
Example 14 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 = {█("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 “she visits A before B“ Hence , E = {█("ABCD, ABDC, ADBC, ACDB, ADBC, ADCB," @"CABD, CADB, CDAB, " @"DABC, DACB, DCAB," )} n(E) = 12 P(E) = (𝑛(𝐸))/(𝑛(𝑆)) = 12/24 = 𝟏/𝟐 Example 14 What is the probability that she visits (ii) A before B and B before C? 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" )} Let F be the event “she visits A before B and B before C “ F = {█("ABCD, ABDC, ADBC , DABC" )} So, n(F) = 4 P(F) = (𝑛(𝐹))/(𝑛(𝑆)) = 4/24 = 𝟏/𝟔 Example 14 What is the probability that she visits (iii) A first and B last? 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" )} Let G be the event “she visits A first and B last” G = {█("ACDB, ADCB" )} So, n(G) = 2 P(G) = (𝑛(𝐺))/(𝑛(𝑆)) = 2/24 = 𝟏/𝟏𝟐 Example 14 What is the probability that she visits (iv) A either first or second? 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" )} Let H be the event “she visits A either first or second” H = {█("ABCD, ABDC, ADBC, ACDB, ADBC, ADCB," @" BACD, BADC,CABD, CADB,DABC, DACB," )} So, n(H) = 12 P(H) = (𝑛(𝐻))/(𝑛(𝑆)) = 12/24 = 𝟏/𝟐 Example 14 What is the probability that she visits (v) A just before B? 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" )} Let I be the event “she visits A just before B” I = {█("ABCD, ABDC, CABD, CDAB, DABC, DCAB," )} So, n(I) = 6 P(I) = (𝑛(𝐼))/(𝑛(𝑆)) = 6/24 = 𝟏/𝟒
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