Question 970612: A man travels to work by car or by train. There is a probability of 2/3 that he travels by train on Monday. if he travels by train on any other day, there
is a probability of 3/4 that he will travel by car the next day. If he travels by car any one day ,there is a probability of 5/6 that he will travel by train the next day. find the probability that he travels:
(a) by car on Monday
(b) by car on Tuesday
(c) by train on Tuesday
(d) by car on Wednesday
(e) by train on Wednesday
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! This might help :) and appreciate your help :) A man travels to work either by car or by train. There is a probability of 2/3 that he travels by train on Monday. if he travels by train on any one day, there is a probability of 3/4 that he will travel by car on the next day. If he travels by car any one day ,there is a probability of 5/6 that he will travel by train the next day. find the probability that he travels: (a) by car on Tuesday (b) by train on Tuesday (c) by car on Wednesday (d) by train on Wednesday ..
T 2/3; :;;C (3/4) ;;;;T(5/6)
T 2/3 ;;;C(3/4) ;;;;; C(1/6)
;;; ;; T (1/4) ;;;; C(3/4)
;;;; T(1/4) ;;;;; T(1/4)
C 1/3;;;;;; C (1/6) ;; C (1/6)
;;;;;;;;;;;;;;;C(1/6) ;; T (5/6)
;;;; T (5/6) ;;;; C (3/4)
;;;;; ;;; T (1/4)
Monday Tuesday Wednesday
Car on Monday =1/3
Car on Tuesday= products of both TC and CC added or (1/2) + (1/6)=2/3
Train on Tuesday= 1/3
Car on Wednesday=(1/9) + (1/4)=13/36
Train on Wednesday=
Products of the three ending in T: (5/12)+(1/24)+(5/108)+(5/72)=(124/216)=(31/54)
Products of the three ending in C: (18/216)+(27/216)+(2/216)+(45/216)=(92/216)=(23/54)
It's tempting to use the last probabilities alone, but for the whole tree diagram, use all of them, because you will have sums greater than 1 and have to deal with the joint probabilities anyway.
Each day has to end up with a probability equal to 1.
This is ambiguous, and I will deal with it literally.
He has a probability of traveling by train on Monday=2/3
He has a 1/3 chance of going by car on Monday. That is A.
The question then says if he travels by train on any other day, there is a probability of 3/4.....
It doesn't say what the probability is that he will travel by train on Tuesday, only that if he does. That is what "any other day" means. The writer of the question may have assumed Monday counted, but any other day does NOT include Monday, and the 2/3 probability applies only to Monday.
I think they want a tree diagram, and this is fine. I can edit the question, but I am not clear what the probability is he will travel by train on Tuesday. Once I know that, I can figure out the rest of the question.
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