Pstat160
Winter 2010
Solution vending machine problem
Part 1.
The can chose the state space to be
S
=
{
0
,
1
,
2
}
with
0
=
no machine down
1
=
one machine down
2
=
two machine down
We denote by
X
n
= # number of machine down, and claim that this define a Markov chain on
S
.
Indeed, this
way the current state encode all the information needed to describe the probability to move to a new state. The
memoryless property of the geometric distribution implies that the probability that a machine is fixed on a given
day (if it has not yet been fixed) is 1
/
5 =
.
2.
Below, we list all possible transitions, and the corresponding event is in
{ }
. Check each time that the probability
addup to 1! Fixing the departing state to be:
i=0:
None of the machines are down.
0
→
0 :=
{
Both machine don’t break
}
, prob=
.
9
2
0
→
1 :=
{
One machine only breaks
}
, prob = 2
×
.
9
×
.
1 =
.
18
0
→
2 :=
{
Both machine break
}
, prob =
.
1
2
=
.
01
i=1:
One machine is down
1
→
0 :=
{
Broken machine fixed, other doesn’t break, prob=
.
2
×
.
9 =
.
18
1
→
1 :=
{
Broken one fixed and other breaks
}
S
{
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 Winter '10
 bonnet
 Probability theory, Exponential distribution, #, Markov chain, π

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