# Probability 2

• February 18th 2006, 04:55 AM
Natasha
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.

• February 18th 2006, 11:30 AM
CaptainBlack
Quote:

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

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

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

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

RonL
• February 18th 2006, 11:33 AM
CaptainBlack
Quote:

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
• February 18th 2006, 11:44 AM
CaptainBlack
Quote:

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:

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

So we may write:

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

$
P(B|late)P(late)=P(B|late)\times 0.325
$

So:

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

RonL
• February 18th 2006, 11:46 AM
Natasha
Thanks Ron!