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mathmath333
 one year ago
Probablity question
mathmath333
 one year ago
Probablity question

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0que: 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?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0The number of ways by which veena can visit is 4!=24

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i can'r go further

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1i guess you can list ways in which a comes before B and count them A bit tedious though

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0yes that is tediuos i want another time saving way

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1or if A is the first then how many ways can the others be visited 3*2 = 6 if A is 2nd and B 3rd or 4th  thats 2 ways

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1 no the last one is 4 ways then we have A 3rd and B 4th  2 ways total 6 + 4 + 2 = 12 so first one is 12/24 = 1/2

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1to be honest I hate probability!

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0sumilarly B before C = 12 ways ?

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1can we multiply these now to get the compound probability?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0ok i need to find now the common between A before B and B before C

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1the question is are they independent events?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i don't understand how independent events means here

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0there are 12 events in both A before B and B before C and i need to find the common events between them

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0(A before B) \(\Large \cap\) (B before C)

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1by listing and counting there are 4 of them

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0ok so i need to list them necessrily

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1i'm afraid i dont remember all the set theory applied to probability

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0ok so same for 3, 4 and 5 , listing ?

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1questions iii to v are quite straightforward  i think!!

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1for q iii it is the number ow ways you can arrange the 3 letters between A and B

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1for iv A first = 3! ways + 31 ways with A second

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1v AB to begin   2 ways AB 2nd and 3rd  2 waya AB last  2 ways

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0AB A_B__ A__B _AB__ _A_B_ _A__B __AB i calculate 7 ways for v.)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439460306887:dw just another way of thinking about it"!

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0is the 7 ways correct for last one ?

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1No i dont think so  isn't it A just before B (A and B together?)

phi
 one year ago
Best ResponseYou've already chosen the best response.1***(i) A before B? *** I guess we could count the different routes but it seems there is nothing special about A and B, and half the time A will be first, and B second (and vice versa)

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1for (v) A just before B? put A and B next to each other and count it as one item let X=AB then we want the number of arrangements of XCD which is 3!=6 the probability easily follows

phi
 one year ago
Best ResponseYou've already chosen the best response.1(ii) A before B and B before C? it seems difficult to find the prob of each event and then subtract off their intersection. Rather, how about writing A B C and asking, where can D go? we can list DABC, ADBC, ABDC, ABCD so 4 patterns out of 24 and prob 4/24= 1/6
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