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que: 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?
The number of ways by which veena can visit is 4!=24
i can'r go further
i guess you can list ways in which a comes before B and count them A bit tedious though
yes that is tediuos i want another time saving way
or 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
- 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
to be honest I hate probability!
sumilarly B before C = 12 ways ?
yes must be
can we multiply these now to get the compound probability?
ok i need to find now the common between A before B and B before C
the question is are they independent events?
i don't understand how independent events means here
there are 12 events in both A before B and B before C and i need to find the common events between them
(A before B) \(\Large \cap\) (B before C)
by listing and counting there are 4 of them
ok so i need to list them necessrily
i'm afraid i dont remember all the set theory applied to probability
ok so same for 3, 4 and 5 , listing ?
questions iii to v are quite straightforward - i think!!
for q iii it is the number ow ways you can arrange the 3 letters between A and B
= 6 right?
3!=6 , yes
for iv A first = 3! ways + 31 ways with A second
v AB to begin - - 2 ways AB 2nd and 3rd - 2 waya AB last - 2 ways
yea it is
AB-- A_B__ A__B _AB__ _A_B_ _A__B __AB i calculate 7 ways for v.)
|dw:1439460306887:dw| just another way of thinking about it"!
is the 7 ways correct for last one ?
No i dont think so - isn't it A just before B (A and B together?)
***(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)
for (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
(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