Martinez
Engr.323
BHW #3
Problem 2.82
Professor Stan der Deviation can take one of two routes on his way home from work. On the first
route, there are four railroad crossings. The probability that he will be stopped by a train at any
particular one of the crossings is 0.1, and train operate independently at the four crossings. The other
route is longer but there are only two crossings, independent of one another, with the same stoppage
probability for each as on the first route. On a particular day, Professor Deviation has a meeting
scheduled at home for a certain time. Which ever route he takes, he calculates that he will be late if he
is stopped by trains at at least half the crossings encountered.
a) Which route should he take to minimize the probability of being late to the meeting?
b) If he tosses a fair coin to decide on a route and he is late, what is the probability that he took the
four-crossing route?
Before starting to go crazy with this problem, lets define some equations that may be helpful for
solving this problem.
An equation that will help us.
P ( Ai1 ∩ ⋅ ⋅ ⋅ ∩ Aik ) = P ( Ai1 ) ∗ P ( Ai2 ) ⋅ ⋅ ⋅ ⋅ ⋅ P ( Aik )
Events A1,…,An are mutually independent if for every k (k = 2,3, …, n) and every subset of indices i1,
i2, . . . , Ik, (pg. 81 Devore)
According to Devore, the events are mutually independent if the probability of the intersection of any
subset of the n events is equal to the product of the individual probabilities.
Definition of some events.
Let:
Ti = {train stops at a railroad crossingi }, for i = 1, 2, …, 6
L = {Professor Stan der Deviation is late }
Note
Ti are mutually independent for i = 1, 2,…,6
Then
Ti = {stopped by train} and Ti’ = {not stopped by train}
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Martinez
Engr.323
BHW #3
Problem 2.82
Therefore
P(Ti) = 0.1 for i = 1, 2, …, 6
Lets visualize the settings for Stan der Deviation preferences for either route. Note that
Ti represent train 1, 2, …, 6
work
Home
Route 1
T1
T2
T3
T4
Route 2
T5
T6
If S. der Deviation takes route # 1 then he’ll have to stop at 1 or fewer stations. Meaning either 0
station or 1 station in order to not be late, so he can stop at zero or 1 stations
Note
Ti = 0.1 for i = 1, 2, …, 6
Ti’ = 0.9
Then solve for the probability of being in time, if he takes route #1.
P(being stopped at 1or fewer stations on route #1) = P(T1 '∩T2 '∩T3 '∩T4 ' ) + P (T1 ∩ T2 '∩T3 '∩T4 ' ) +
P(T1 ∩ T2 ∩ T '3 ∩T ' 4 ) + P(T1 '∩T2 '∩T3 ∩ T4 ' ) + P(T1 '∩T2 '∩T3 '∩T4 )
'
Note there four ways to be stopped y exactly one train in Route #1.
Then plugging the values for (Ti) and (Ti’)
2
Martinez
Engr.323
BHW #3
Problem 2.82
P(being stopped at 1 or fewer stations on route #1) = .94 + 4(0.1) (.9)3
P(being stopped at 1 or fewer stations) = 0.9477
P(being late if he takes Route #1) = 1 – 0.9477 = 0.0523
If the Professor takes route #2 use the basically the same approach. Remember he can not stop at any
station, otherwise he’ll be late. Therefore we need to use Ti’
P( stopped at 0 stations ) = (T5 '∩T6 ' )
P( stopped at 0 stations) = [(0.9 ) ∗ (0.9 )]
P ( stopped at 0 stations) = 0.81
P(being late if he takes Route # 2) = 1 – 0.81 = 0.19
This means that S. der Deviation should take route #1
b.
Then let’s apply conditional probability
Let
RT1 = {took four railroad crossings trip}
L = {Late}
½ chance he took Route #1
½ chance he took Route #2
Then
P( RT 1 | L ∩ lossed fair coin) =
P( RT 1 ∩ L ∩ fair coin)
P( L ∩ fair coin)
P( RT 1 | L ∩ lossed fair coin) =
0.5 * (1 − 0.9477)
0.5 * (1 − 0.9477) + 0.5 * (1 − 0.81)
P( RT | L ∩ lossed fair coin) = 0.215848
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