#4 HOMEWORK SOLUTIONS 1.
0 1 2
0.4 0.3 0.3
0
0 0 0 1
0.2 0.7 01 .
2
1 1 1
0.8 0.0 0.2
2 2 0 2
0.4,
1
0.4,
2
0.2
1
2. a)
20
1 0.2
20
0.0
10
20
125 .
b)
P( Going to state 0 after at least two steps)
3.
1
1 f 20
1
0.8
0.2
P
a)
0.9 0.1 0.12 0.88
P2
0.822 0.178 0.2136 0.7864
P3
0.7612 0.2388 0.2866 0.7134
P(x1=A) = (0.65)*(0.9)+(0.35)*(0.12) = 0.627
b)
P(x3=A) = (0.65)*(0.7612)+(0.35)*(0.2866) = 0.595
c)
A C
0.9 0.1 1
A A A
0.12 0.88
C
C C A
0.545
d)
( f 013) (3 p 01) (1 (2 f 01 ) p11 ) ( f 012) p11
0.2388 0.1 * 0.7864 0.09 * 0.88 0.081
4. a) let 0=operational, 1=down, 2=repaired
0.9 0.1 0
then P
1 , we also have 0.9 0.1 0
0
0
ij
1
k 20
1 j
pik
kj
00
So,
10 0.1 10 1 20 1 0.1
10
00
such that
10 20
11/ 9 20 / 9 11/ 9
01
01 11 21
1 0.9 01 1 21 1 0.9 1 0.9
12 01
such that
11 21
10 11 10 11 1 11
21
02
02
0.1 0.1
12
00
1 0
02 12
such that
10 20
22
1 0.9
The expected number of days that elevator will remain operational is b) It would be the same due to the Markovian property.
=10.
c) P 3
0.819 0.091 0.09 0.81 0.09 0.1 0.819 0.091 0.09
The probability that the elevator will be operational at the beginning of the third (3 day given that the elevator is currently operational = p 00) = 0.819. d)
( f 013) (3 p 01) ( (2 f 011) p11 ) ( f 012 ) p11
0.091 0.1 * 0.1 0.09 * 0 0.081