| Temperature | Skin | Blood Pressure | Blocked Nose | Diagnosis | | ----------- | ------ | -------------- | ------------ | --------- | | Low | Pale | Normal | True | N | | Moderate | Pale | Normal | True | B | | High | Normal | High | False | N | | Moderate | PaleFF | Normal | False | B | | High | Red | High | False | N | | High | Red | High | True | N | | Moderate | Red | High | False | B | | Low | Normal | High | False | B | | Low | Pale | Normal | False | B | | Low | Normal | Normal | False | B | | High | Normal | Normal | True | B | | Moderate | Normal | High | True | B | | Moderate | Red | Normal | False | B | | Low | Normal | High | True | N | | | Temperature | | | Skin | | | Pressure | | | Blocked | | Diag | | | -------- | ----------- | --- | ------ | ---- | --- | ------ | -------- | --- | ----- | ------- | --- | ---- | ---- | | | N | B | | N | B | | N | B | | N | B | N | B | | Low | 2 | 3 | Pale | 1 | 3 | Normal | 1 | 6 | True | 3 | 3 | 5 | 9 | | Moderate | 0 | 5 | Normal | 2 | 4 | High | 4 | 3 | False | 2 | 6 | | | | High | 3 | 1 | Red | 2 | 2 | | | | | | | | | | | Temperature | | | Skin | | | Pressure | | | Blocked | | | Diag | | Low | 2/5 | 3/9 | Pale | 1/5 | 3/9 | Normal | 1/5 | 6/9 | True | 3/5 | 3/9 | 5/14 | 9/14 | | Moderate | 0/5 | 5/9 | Normal | 2/5 | 4/9 | High | 4/5 | 3/9 | False | 2/5 | 6/9 | | | | High | 2/5 | 1/9 | Red | 3/5 | 2/9 | | | | | | | | | # Problem 1 # $Pr[Diagnosis=N|E] = \frac{2}{5} \times \frac{2}{5} \times \frac{4}{5} \times \frac{3}{5} \times \frac{5}{14} = 0.027428571$ # $Pr[Diagnosis = B|E] = \frac{3}{9} \times \frac{4}{9} \times \frac{3}{9} \times \frac{3}{9} \times \frac{9}{14} = 0.010582011$ # $p(B) = \frac{0.0106}{0.0106+0.0274} = 0.2789$ # $p(N) = \frac{0.0274}{0.0106+0.0274} = 0.7211$ Diagnosis N is much more likely than Diagnosis B # Problem 2 # $Pr[Diagnosis = N|E] = \frac{2}{5} \times \frac{1}{5} \times \frac{3}{5} \times \frac{5}{14} = 0.0171$ # $Pr[Diagnosis = B|E] = \frac{3}{9} \times \frac{6}{9} \times \frac{3}{9} \times \frac{9}{14} = 0.0476$ # $p(N) = \frac{0.0171}{0.0171+0.0476} = 0.2643$ # $p(B) = \frac{0.0474}{0.0476+0.0171} = 0.7357$ Diagnosis B is much more likely than Diagnosis N # Problem 3 # $Pr[Diagnosis = N|E] = \frac{0}{5} \times \frac{2}{5} \times \frac{4}{5} \times \frac{3}{5} \times \frac{5}{14} = 0$ # $Pr[Diagnosis = B|E] = \frac{5}{9} \times \frac{4}{9} \times \frac{3}{9} \times \frac{3}{9} \times \frac{9}{14} = 0.018$