A visitor is coming for a seminar in UQ. The probability that he comes by train, car and flight is respectively 0.5, 0.3 and 0.2. And the probability of delay by these three means of transport is respectively 0.1, 0.3 and 0.4. If the visitor is late, what is the probability that he came by train? (The result retains four decimals)
Question
A visitor is coming for a seminar in UQ. The probability that he comes by train, car and flight is respectively 0.5, 0.3 and 0.2. And the probability of delay by these three means of transport is respectively 0.1, 0.3 and 0.4. If the visitor is late, what is the probability that he came by train? (The result retains four decimals)
Solution
This is a problem of conditional probability and can be solved using Bayes' theorem. Bayes' theorem is a way to find a probability when we know certain other probabilities.
The formula for Bayes' theorem is:
P(A|B) = [P(B|A) * P(A)] / P(B)
where:
- P(A|B) is the probability of event A given event B is true
- P(B|A) is the probability of event B given event A is true
- P(A) and P(B) are the probabilities of events A and B
In this case, we want to find the probability that the visitor came by train given that he is late. So, event A is "came by train" and event B is "is late".
We know that:
- P(A) = P(train) = 0.5
- P(B|A) = P(late|train) = 0.1
We also need to find P(B), the probability that the visitor is late. This is the sum of the probabilities of being late for each mode of transport, each multiplied by the probability of choosing that mode of transport:
P(B) = P(late) = P(late|train)P(train) + P(late|car)P(car) + P(late|flight)P(flight) = (0.10.5) + (0.30.3) + (0.40.2) = 0.05 + 0.09 + 0.08 = 0.22
Now we can plug these values into Bayes' theorem:
P(A|B) = [P(B|A) * P(A)] / P(B) = (0.1 * 0.5) / 0.22 = 0.2273 (rounded to four decimal places)
So, the probability that the visitor came by train given that he is late is 0.2273.
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