Question 1202957: A man travel to work either by car or by bus. If he goes by car the probability that he will be late is 1/5, but if he goes by bus the probability is 1/4.
(a) if he travels by car every day for four successive days, what is the probability that he will be late only once?
(b) if he is twice likely to travel by bus as by car, what is the probability that he will be late on a particular day.
What I have tried.
a)P(late by car)=1/5
P(late by bus)=1/4
P(late once)= P(on time for 3 days)
Let the days on time be: A,B,C,D
Let the day of late be: À, ß, Ç, d
Therefore,
P(on time for 3 days)= (A×B×C×d)+(A×B×Ç×D)+(A×ß×C×D)+(À×B×C×D)
P(on time for 3 days)= ((4/5)³×1/5)×4
P(on time for 3 days)= (Probability of late once)= 256/625.
b) let
A,B= days traveled by car
C,D= days traveled by bus
P(late on a particular day)= (A×B×C×d)+(A×B×Ç×D)+(A×ß×C×D)+(À×B×C×D)
P(late on a particular day)= ((1/5)²×(3/4×1/4))×2+((1/4)²×4/5×1/5)×2
P(late on a particular day)= 7/200.
However, I'm not sure of my answer.
Answer by greenestamps(13195)   (Show Source):
You can put this solution on YOUR website!
Thanks for showing your work. While the instructions for posting a question say you should do that, it is extremely uncommon for someone to do that.
Your work for the first question is fine:
P(late 1 of 4 days traveling by car) = (C(4,1))*((4/5)^3)*((1/5)^1) = 256/625
I don't understand your setup for the second problem....
He is twice as likely to travel by bus as by car, so:
P(travel by bus) = 2/3
P(travel by car) = 1/3
P(travel by bus AND is late) = (2/3)(1/4) = 1/6
P(travel by car AND is late) = (1/3)(1/5) = 1/15
P(late) = 1/6 + 1/15 = 5/30 + 2/30 = 7/30
ANSWER: 7/30
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Added after question from student....
The "3" has nothing to do with a sample space.
Because he is twice as likely to travel by bus as by car, the probability that he travels by bus is 2/3 and the probability that he travels by car is 1/3.
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