Probablity question

- mathmath333

Probablity question

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- mathmath333

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?

- anonymous

probably not

- mathmath333

The number of ways by which veena can visit is 4!=24

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## More answers

- welshfella

yes

- mathmath333

i can'r go further

- mathmath333

*can't

- welshfella

i guess you can list ways in which a comes before B and count them A bit tedious though

- mathmath333

yes that is tediuos i want another time saving way

- welshfella

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

- welshfella

- 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

to be honest I hate probability!

- mathmath333

sumilarly B before C = 12 ways ?

- welshfella

yes must be

- welshfella

can we multiply these now to get the compound probability?

- mathmath333

ok i need to find now the common between A before B and B before C

- welshfella

the question is are they independent events?

- mathmath333

i don't understand how independent events means here

- mathmath333

there are 12 events in both A before B and B before C and i need to find the common events between them

- mathmath333

(A before B) \(\Large \cap\) (B before C)

- welshfella

by listing and counting there are 4 of them

- mathmath333

ok so i need to list them necessrily

- welshfella

i'm afraid i dont remember all the set theory applied to probability

- mathmath333

ok so same for 3, 4 and 5 , listing ?

- welshfella

questions iii to v are quite straightforward - i think!!

- welshfella

for q iii it is the number ow ways you can arrange the 3 letters between A and B

- welshfella

= 6 right?

- mathmath333

3!=6 , yes

- welshfella

for iv
A first = 3! ways
+ 31 ways with A second

- welshfella

* 3!

- mathmath333

ok

- welshfella

v
AB to begin - - 2 ways
AB 2nd and 3rd - 2 waya
AB last - 2 ways

- anonymous

nice question

- welshfella

yea it is

- mathmath333

AB--
A_B__
A__B
_AB__
_A_B_
_A__B
__AB
i calculate 7 ways for v.)

- mathmath333

@welshfella

- anonymous

|dw:1439460306887:dw| just another way of thinking about it"!

- mathmath333

is the 7 ways correct for last one ?

- welshfella

No i dont think so - isn't it A just before B (A and B together?)

- phi

***(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

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

- phi

(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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