SOLUTION: 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

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Question 970643: 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 ..
Answer by mathmate(429) About Me  (Show Source):
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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 ..

Properly capitalizing the text reduces the chance of errors of interpretation.

Given:

Probabilities of travelling by car (P(C)=2%2F3) or train (P(T)=1-2%2F3=1%2F3) on Monday.
Probabilities of each mode of transportation based on outcome of previous day:
- P(CT)=3%2F4 => P(CC)=1-3%2F4=1%2F4
- P(TC)=5%2F6 => P(TT)=1-5%2F6=1%2F6
Need to draw a tree diagram of all combinations for three days (Monday to Wednesday) from which conclusions may be drawn.

Solution:
Since it is a three step process, we have a three level tree-diagram with a total of 2 outcomes for Monday {C,T}, 4 outcomes for Tuesday {CC,CT,TC,TT}, and eight outcomes for Wednesday {CCC,CCT,CTC,CTT,TCC,TCT,TTC,TTT}.
The probabilities of outcomes of each day are shown in shown in separate columns, the sum of which must equal 1 or else there is a mistake somewhere.

The tree diagram is shown below:


We can read off the answers from the tree diagram as follows:
X means any outcome, T=by train, C=by car
(a) by car on Tuesday
P%28XC%29=1%2F2%2B1%2F18=10%2F18=5%2F9

(b) by train on Tuesday
P(XT)=1%2F6%2B5%2F18=8%2F18=4%2F9
or
P(XT)=1-P%28XC%29=1-5%2F9=4%2F9
(c) by car on Wednesday
P(XXC)=1%2F8%2B1%2F12%2B5%2F24%2B1%2F108=23%2F54
P(XXT)=1-23%2F54=31%2F54