1. ## Probability 2

A town has three bus routes A, B and C. Route A has twice as many buses as each of B and C. Over a perido of time it has been found that, along a certain stretch of road, where the three routes converge, the buses on these routes run more than five minutes late 1/2, 1/5 and 1/10 of the time respectively.

1) A bus is going down this stretch of road. What is the probability it is more than 5 minutes late?

2) Comment on the size of the answer to 1) with the respect to the given probabilities 1/2, 1/5 and 1/10.

3) An inspector, standing on this particular stretch of road, sees a bus that is more than five minutes late. Find the probability that it is a route B bus.

2. Originally Posted by Natasha
A town has three bus routes A, B and C. Route A has twice as many buses as each of B and C. Over a perido of time it has been found that, along a certain stretch of road, where the three routes converge, the buses on these routes run more than five minutes late 1/2, 1/5 and 1/10 of the time respectively.

1) A bus is going down this stretch of road. What is the probability it is more than 5 minutes late?
Part 1)

There are twice as many rout $\displaystyle A$ buses as either route $\displaystyle B$ or $\displaystyle C$, so half of the
buses are route $\displaystyle A$. So the proportion $\displaystyle P(A)$ of buses which are route $\displaystyle A$ is $\displaystyle 0.5$, and
the proportion$\displaystyle P(B)$ that are route $\displaystyle B$ is 0.25, and the proportion $\displaystyle P(C)$ route C
is again 0.25. So the probabilty that a random bus is late is:

$\displaystyle P(late)=P(A)P(late|A) + P(B)P(late|B)+P(C)P(late|C)$

But we are told that the proportion $\displaystyle P(late|A)$ of route $\displaystyle A$ buses which are late is $\displaystyle 1/2=0.5$,
and that $\displaystyle P(late|B)=1/5=0.2$, and $\displaystyle P(late|C)=1/10=0.1$. So:

$\displaystyle P(late)=0.5 \times 0.5 + 0.25 \times 0.2+0.25 \times 0.1=0.325$

RonL

3. Originally Posted by Natasha
A town has three bus routes A, B and C. Route A has twice as many buses as each of B and C. Over a perido of time it has been found that, along a certain stretch of road, where the three routes converge, the buses on these routes run more than five minutes late 1/2, 1/5 and 1/10 of the time respectively.

1) A bus is going down this stretch of road. What is the probability it is more than 5 minutes late?

2) Comment on the size of the answer to 1) with the respect to the given probabilities 1/2, 1/5 and 1/10.
Part 2)

Answer to 1) is 0.325, which is less than the proportion of route A buses
which are late because the lateness of route A is diluted by the better
time keepers of routs B and C, but only slightly because half of all buses
are the poor time keepers of route A.

RonL

4. Originally Posted by Natasha
A town has three bus routes A, B and C. Route A has twice as many buses as each of B and C. Over a perido of time it has been found that, along a certain stretch of road, where the three routes converge, the buses on these routes run more than five minutes late 1/2, 1/5 and 1/10 of the time respectively.

1) A bus is going down this stretch of road. What is the probability it is more than 5 minutes late?

2) Comment on the size of the answer to 1) with the respect to the given probabilities 1/2, 1/5 and 1/10.

3) An inspector, standing on this particular stretch of road, sees a bus that is more than five minutes late. Find the probability that it is a route B bus.
Part (3).

This is simply an aplication of Bayes theorem:

$\displaystyle P(B \wedge late)=P(B)P(late|B)=P(B|late)P(late)$

So we may write:

$\displaystyle P(B)P(late|B)=0.25 \times 0.2$
$\displaystyle P(B|late)P(late)=P(B|late)\times 0.325$

So:

$\displaystyle P(B|late)=\frac{0.25 \times 0.2}{0.325}=0.154$

RonL

5. Thanks Ron!