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Assume that there is a choice of three railway stations. The probability of either walking or driving to each station is known. This is specific to each station. The probability of choosing each of the stations given walk or drive is also known. What I require is the overall probability of each of the three stations being chosen (summing to 1). Example:

Station A

  • Probability of walking to it 0.8
  • Probability of driving to it 0.2
  • Probability of choosing this station if walk 0.5
  • Probability of choosing this station if drive 0.3

Station B

  • Probability of walking to it 0.1
  • Probability of driving to it 0.9
  • Probability of choosing this station if walk 0.2
  • Probability of choosing this station if drive 0.2

Station C

  • Probability of walking to it 0.5
  • Probability of driving to it 0.5
  • Probability of choosing this station if walk 0.3
  • Probability of choosing this station if drive 0.5

The probabilities of choosing the stations necessarily sum to one for walk or drive. I'm look for a result like:

  • Probability of station A: x
  • Probability of station B: y
  • Probability of station C: z

Where x, y , z sums to 1.

Thanks for any help. Not even sure this is possible.

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  • $\begingroup$ Before looking at the algebraic terms x,y and z why don't the known probabilities sum to 1 in each case. $\endgroup$ – Michael R. Chernick Mar 7 '17 at 21:04
  • $\begingroup$ Define your events: W - walking, D - driving, A - choosing station A, B - choosing station B, C - choosing station C. Now state all your probabilities in terms of the events A,B,C,W,D. All of them will be conditional probabilities. Use the conditional probability formula and see if you can solve for p(A) p(B) p(C).. $\endgroup$ – Zahava Kor Mar 7 '17 at 21:07
  • $\begingroup$ Hi Michael, do you mean the probabilities of walking or driving to the station? This probability is specific to each station, so it sums to one for each station but not across stations. The proportion walking or driving depends on the distance to the station which is station specific. $\endgroup$ – Matty1965 Mar 7 '17 at 21:12
  • $\begingroup$ This is definitely not from a text book. It is a problem that I am trying to resolve for my research. $\endgroup$ – Matty1965 Mar 7 '17 at 21:15
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    $\begingroup$ How is this supposed to work? Is there a draw for walk vs drive, & then a draw of which station? How are the probabilities you list supposed to go together? $\endgroup$ – gung - Reinstate Monica Mar 7 '17 at 22:07
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By defining the events W,D,A,B,C as I wrote in the comment, your first 4 probabilities are all conditional probabilities: P(W/A)=0.8 ; P(D/A)=0.2 ; P(A/D)=0.5 ; P(A/W)=0.3. You can go on like that and write the conditional probabilities for the other two stations. You get 12 conditional probabilities. Write each of them in the form P(A/B) = P(A∩B)/P(B) and try to solve for P(A),P(B),P(C).

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  • $\begingroup$ Hi, thanks for your help. The "try and solve it" will probably be the sticking point as I'm not an expert in stats, which is why I was asking for a solution here. I can do fairly simple stuff, but this goes beyond my knowledge. $\endgroup$ – Matty1965 Mar 7 '17 at 22:52

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