ECMT 3160: Example 2.9.5: Snow
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ECMT 3160: Example 2.9.5: Snow
2022
There are 5 roads which we can label as follows: R1 represents the road A to B ; R2 represents the road B to D ; R3 represents the road A to C; R4 represents the road C to D ; and R5 represents the road B to C. In this notation, the outcomes of this experiment is the set of all ordered tuples with 5 components given by
Ω = ((R1, R2 , R3 , R4 , R5 ) : R; e (0, 1}, i = 1, 2, 3, 4, 5}.
In this notation the event (R; = 1} means that the corresponding road is open, and (R; = 0} means that it is closed. We are also given P [R; = 0] = σ for each i, and these events are independent.
The example proceeds using the partition rule by conditioning on the event R = (road BC is open}. In my notation, this event is (R5 = 1}. Also, recall that Q is the event that you can drive from A to D. To answer the question posed on the Discussion Board, we need to figure out the probabilities of the two events: (Q n Rc } and (Q n R}. This is because
Q = (Q n Rc) u (Q n R) ,
which is the union of disjoint events, is the basis for the use of the Partition Rule.
From the above notation we have that
(Q n Rc} = ((R1 = 1} n (R2 = 1}) u ((R3 = 1} n (R4 = 1}) n (R5 = 0}
= (((R1 = 1} n (R2 = 1})c n ((R3 = 1} n (R4 = 1})c)c n (R5 = 0} and (Q n R} = ((R1 = 1} u (R3 = 1}) n ((R2 = 1} u (R4 = 1}) n (R5 = 1}.
A side note: think in words what these decompositions of the events (Q n Rc } and (Q n R} is stating. Now we are in a position to apply the Rules of Probability to compute the probabilities of these event.
We first focus on the event (Q n Rc }. Because this is characterized as an intersection of other events, which are independent events, we can use the Product Rule to calculate P [Q n Rc]. But it is not so straight- forward as you will also need the Complement Rule. In particular,
P [Q n Rc] = (1 - P [((R1 = 1} n (R2 = 1})c] P [((R3 = 1} n (R4 = 1})c]) P [R5 = 0] = ╱ 1 - ╱ 1 - (1 - σ)2、╱ 1 - (1 - σ)2、、σ = ╱ 1 - ╱ 1 - (1 - σ)2、2、σ.
Now using the definition of conditional probability, observe that
P [Q | Rc] = = 1 - ╱ 1 - (1 - σ)2、2 ,
as required since we know that P [Rc] = σ .
Next, we focus on P [Q n R]. Using the result of Exercise 11 in Section 2.16 of the book, we obtain
((R1 = 1} u (R3 = 1}) = [((R1 = 1} u (R3 = 1})c]c = [((R1 = 1}c n (R3 = 1}c)]c ((R2 = 1} u (R4 = 1}) = [((R2 = 1} u (R4 = 1})c]c = [((R2 = 1}c n (R4 = 1}c)]c .
The above is advantageous because we have now expressed these events in the form of the complement of intersections of events. As they are independent of (R5 = 1}, we can firstly calculate their probability separately and then multiply it with P [R5 = 1] to obtain P [Q n R]. Towards that end, by the Complement
Rule:
P [(R1 = 1} u (R3 = 1}] = 1 - P [((R1 = 1}c n (R3 = 1}c)] = 1 - P [((R1 = 0} n (R3 = 0})] P [(R2 = 1} u (R4 = 1}] = 1 - P [((R2 = 1}c n (R4 = 1}c)] = 1 - P [((R2 = 0} n (R4 = 0})] .
Now we are in a position to apply the Product Rule to simplify the above expressions:
P [(R1 = 1} u (R3 = 1}] = 1 - P [((R1 = 0} n (R3 = 0})] = 1 - P [R1 = 0] P [R3 = 0] = 1 - σ 2 P [(R2 = 1} u (R4 = 1}] = 1 - P [((R2 = 0} n (R4 = 0})] = 1 - P [R2 = 0] P [R4 = 0] = 1 - σ 2
Consequently, we use the Product Rule again to obtain
P [Q n R] = ╱ 1 - σ 2、2 P [R5 = 1] = ╱ 1 - σ 2、2 (1 - σ) .
Therefore,
P [Q | R] = = ╱ 1 - σ 2、2 ,
by the definition of Conditional Probability.
2022-04-06