Question 1 (10 points): Let R be the event that the chosen ball is red. Let Bi be the event that a person chooses Basket i.

We already know that: P(R|B1) = 0.75 P(R|B2) = 0.60 P(R|B3) = 0.45

The chance that the person choose the three baskets are same. The person now choose one of the baskets at random and then pick a ball from the chosen basket, also at random. What is the probability that the chosen ball is red?" P(R|B1) 0.75 P(R|B2) 0.6 P(R|B3) 0.45 P(B1)=P(B2)=P(B3) 0.33 Your answer:

Question 2 (20 points): Suppose {A, B} is conditional independent on C, and {A, B} is also conditional independent on C' (Pr(C)+Pr(C')=1). We already know:

P(A|C) = 0.4, P(B|C) = 0.6, P(A|C') = 0.3, P(B|C') = 0.2

Show wether or not A and B is independent.

Hint: P(AB) = P(A|C)P(B|C)P(C) + P(A|C')P(B|C')P(C') P(C) 0.7 Q2.1: What is P(A)? P(C') 0.3 Q2.2: What is P(B)? P(A|C) 0.4 Q2.3: What is P(AB)? P(B|C) 0.6 Q2.4: What is P(A)*P(B)? P(A|C') 0.3 Q2.5: Whether A and B are independent? P(B|C') 0.2

Question 3 (10 points): It is known that 50% of Chrome add-ons are spam add-ons containing virus. A tool has been developed to detect whether an add-on is spam or not. It claims that it can detect 99% of spam, but it can still make mistakes: a non-spam detected as spam is 5%.

We are known that an add-on is detected as spam, what's the probablity that it is in fact a non-spam add-on P(B'|A)?

A = event that an add-on is detected as spam B = event that an add-on is spam B' = event that an add-on is not spam" P(B) 0.5 P(B') 0.5 P(A|B) 0.99 P(A|B') 0.05 Your answer:

Question 4 (10 points): A clinician is asked to see a sick person. The clinician has acknowledged that 90% of sick people in the hospital have the fever (F), while the remaining 10% are sick with hypertension (H).

A well-known symptom of hypertension is irregular heart rhythms (denoted as R). Assuming the probablity of having an irregular heart rhythms if the person has hypertension is P(R|H) = 0.95.

However, people with fever also may show irregular heart rhythms, and the probablity is P(R|F) = 0.08.

After checking the person, the clinician confirmed an irregular heart rhythms. What's the probablity that the person has hypertension P(H|R)? " P(F) 0.9 P(H) 0.1 P(R|H) 0.95 P(R|F) 0.08 Your answer: