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Summary CPSC 322 Assignment 4 Solutions
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Solutions for Assignment 4 - CPSC 322
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Question 1 Probabilities(a)
Answer: 0.4 should be 0
(b)
Answer: [.5, .5] , [.5, .5]; they are not independent eg. P(A=T,B=T)=.2 which is different from
.25, .3 / .5 = .6
(c)
Yes, because they are both the result of applying the chain rule to the same joint but with
different orderings for the variables.
, Question 2 Bayes' rule
(a)
Answer: Let W = 1 denote a warning for a failure in the electrical system and W = 0 denote no warning. Likewise,
let F = 1 denote that your car has a failure in its electrical system that will cause a fire and F = 0 denote that your
car does not have this failure. We know
P(F = 1) = 1/10000 = 0.0001 (1)
P(W = 1 | F = 1) = 0.95 (2)
P(W = 0 | F = 0) = 0.95 (3)
First, we apply the rule of total probability to compute P(W = 1):
P(W = 1) = P(W = 1 | F = 0) ∙ P(F = 0) + P(W = 1 | F = 1) ∙ P(F =1) (4)
= 0.05 ∙ 0.9999 + 0.95 ∙ 0.0001 (5)
≈ 0.0501 (6)
We want to know P(F = 1 | W = 1). Applying Bayes' rule, we have
P(W = 1|F = 1)∙ P(F = 1)
P(F = 1 | W = 1) = (7)
P(W = 1)
0.95 ∙ 0.0001
= (8)
P(W = 1)
≈ 0.0019 (9)
(b)
Answer:
P(F = 1) = 1/100 = 0.01 (1)
P(W = 1 | F = 1) = 0.95 (2)
P(W = 0 | F = 0) = 0.95 (3)
First, we apply the rule of total probability to compute P(W = 1):
P(W = 1) = P(W = 1 | F = 0) ∙ P(F = 0) + P(W = 1 | F = 1) ∙ P(F =1) (4)
= 0.05 ∙ 0.99 + 0.95 ∙ 0.01 (5)
≈ 0.059 (6)
We want to know P(F = 1 | W = 1). Applying Bayes' rule, we have
P(W = 1|F = 1)∙ P(F = 1)
P(F = 1 | W = 1) = (7)
P(W = 1)
0.95 ∙ 0.01
= (8)
P(W = 1)
≈ 0.16 (9)
(c)
Answer: Because this is the prior probability of your car having the failure, which directly (and proportionally)
influences the posterior. So, even though the self-diagnostic test is quite accurate, it is still imprecise enough for it
to remain unlikely that your car will experience a fire. If you are surprised by the outcome you are not alone:
humans are typically not good at this type of probabilistic reasoning and believe a 95% precise test more than they
ought to.
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