Help me solve a question on probability theory

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Yael

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probability visit denver before san francisco

Hi everyone

I'm stuck with this question of probability theory. the first part is alright, the second part - i know the answer but can't seem to use the formulas to show it

"A businesswoman in philadelphia is preparing an itinerary for a visit to six major cities. the distance traveled, and hence the cost of the trip, will depend on the order in which she plans her route.

a. how many different itineraries (and trip costs) are possible?
b. if the businesswoman randomly selects on of the possible itineraries and denver and san francisco are two of the cities that she plans to visit, what is the probability that she will visit denver before san francisco:?:

any help/ suggestions would be much apreciated
 

a businesswoman in philadelphia is preparing

My engineering instincts say 1/2 as the answer. You might want to make an exhaustive list (all 6! of entries) and count. This might be done on a computer program.
 

a businesswoman in philadelphia

that's what i thought. 50%.
the only thing is i somehow have to use probability's formulas to proove it (it's a course assignment i need to post). so i don't think the computer program or drawing will work :-/
 

Re: probability question

Try this method. There is a 1/6 chance that the first city of the six is Denver. San Francisco must be a latter one.

Then there is a 4/6 chance that the first city is neither x a 1/5 chance that Denver is the second one visited and San Francisco is visited later.

Then there is a 4/6 chance that the first is neither x a 3/5 chance that the second city is neither x a 1/4 chance that the third city visited is Denver and San Francisco is visited later.

Continue this for the first five cities visited. Add the total.
 

Re: probability question

Hello,
I did the calculations and goes as follow:
there are 6! possibilities in total.
However, there are 5! possibilities for denver to be before san francisco.

if u denote them as : Denver = 1
San Francisco = 2

1 2 x x x x
1 x 2 x x x
1 x x 2 x x 5 possibilities but u need to multiply by 24 since
1 x x x 2 x 3 4 5 6 can be arranged in 4! ways that is 24
1 x x x x 2 so total for this arrangment is 24 X 5 = 120

x 1 2 x x x
x 1 x 2 x x
x 1 x x 2 x Same here 120
x 1 x x x 2
x 1 x x x x

and you go on so u need to multiply 120 by 5 to check the possibilities of denver being before san francisco 120 X 5 = 600
However, The total number of possibilities of the cities visits is 6! which is 720
Therefore the probability of denver being before San Francisco = 600/720 = 0.83333 .

I hope that the answer consist with the expect results of yours.
 

Re: probability question

i would answer:

since she visits them one after the other, there are two possibilities:
1. D before S
2. S before D

these are the events.

one of these possibilites will certanly occur, and since they exclude ( P(1. and 2.)=0) each other you can say:
P(D before S)+P(S before D)=1 (1)

so the union of the events is the certain event

but now there is no reason, why we should prefer the event DbeforeS or SbeforeD,
S is as good as D
so:
P(D before S)=P(S before D)

and together with (1): P(D before S)=P(S before D)=1/2
 

probability question

If she doesnt have to pick 2 but any number of cities from 1 to 6, then there are 1p6 + 2p6p + 3p6+ 4p6 + 5p6 + 6p6 possibilities, because there are 6 possible numbers of visits she can do in one trip and for each one order matters as well.

For b, the answer is 1/2 I believe. You can show this as number of times the city comes up / number of possible cities, this is just 1/2.
 

Re: probability question

You should have been suspicious of your answer since the probablity for the reverse question (SF before D) must have the same probabilty and the two added together would be more than 1.0.

The mistake you made is in your diagrams or maybe assumption:
The x1xxxx is invalid since it does not contain 2. She must visit both cities. So the block totals are
12xxxx 24x5
x12xxx 24x4
xx12xx 24x3
xxx12x 24x2
xxxx12 24x1
---------------
total 360
prob = 360/720 = 0.5
 

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