# Solutions 3

### Exercise 2.3:

#### Part i:

See the Hugin network.

#### Part ii:

P(Inf|Test=pos)=(0.065,0.935)

### Exercise 2.4:

#### Part i:

See the Hugin network.

#### Part ii:

P(Infected|Test=pos)=(0.546,0.454)

### Exercise 2.6:

#### Part i:

P(Pr|Bt=n,Ut=n)= (0.53,0.47), see also the Hugin network.

#### Part ii:

P(BT|Pr=y)=(0.64,0.36)
P(BT|Pr=n)=(0.106,0.894)
P(UT|Pr=y)=(0.73,0.27)
P(UT|Pr=n)=(0.107,0.893)

P(Pr|Bt=n,Ut=n)= (0.449,0.551), see also the Hugin network.

### Exercise 2.7:

P(Bt=y,Ut=y)= (0.441365).

### Exercise 2.11:

#### Part i:

See the Hugin network.

#### Part ii:

See the Hugin network.

#### Part iii:

Use the Hugin network and try out the different configurations. The most probable symbols will (for this particular case) correspond to the most probable word, namely baaaa. However, notice the difference in their probabilities.

### Exercise 2.13:

Yes, see the See the Hugin network.

### Exercise 2.16:

See the Hugin network.

### Exercise 2.17:

A solution to the exercise can be found here in either postscript or PDF.

### Exercise 2.18:

#### Part i:

The proof follows the proof of Exercise 2.17.

#### Part ii:

To specify the or-gate in Exercise 2.16 we need a table with 64 entries. However, for the other two models we can represent the or-gate using tables with a total of 32 entries.

### Exercise 2.19:

#### Part i:

See the Hugin network.

#### Part ii:

See the Hugin network.

### Exercise 2.21:

#### Part i:

See the Hugin network. As the probability of Result=yes is positive, there are assignments of truth values making the expression true.

#### Part ii:

Insert A=n and B=y as evidence and propagate. As the probability of Result=yes is positive, there are assignments of truth values to the remaining variables making the expression true. If you insert Result=y propagate, you will see that the assignment to the remaining variables must be C=y, D=y, E=y and F=y.