[Alice, Bob, and Chess] Busses are often late. Bus 5A Amber and bus 9B Brown are
ID: 1808253 • Letter: #
Question
[Alice, Bob, and Chess] Busses are often late. Bus 5A Amber and bus 9B Brown are late with probabilities P(A) =0.1 and P(B) = 0.25, respectively. A reporter from the Daily Illini has observed that, whenever bus 5A is late, bus 9B is also late, i.e., P(B|A) = 1.0. The reporter suspects that Bob, the driver of bus 9B, and Alice, the driver of bus 5A, are secretly meeting to play chess once every few days, and that their chess games make them late. Alice defends herself by saying that P(AB) is only 1/10; since P(AB) = P(A), she argues, the events A and B must be independent. What's wrong with her argument? The reporter wants to find out how often Bob and Alice play chess. Let C be the event "Bob and Alice play chess." The reporter postulates that P(A|C) = 1 and P(B|C) = 1, i.e., a chess game always makes them late. She further postulates that, on days without chess, Bob and Alice are independent, i.e., P(AB|C C) = P(A|C C)P(B|C C). Under these assumptions (including the assumption P(AB) = 0.1 from part (a)), what is P(C)?Explanation / Answer
a) Her argument is wrong as A and B are independent if and only if P(AB)= P(A)*P(B).. but here P(AB)= P(A)= 1/10 while P(A)*P(B)= 0.1* 0.25= 0.025
Hence P(AB) is not equal to P(A)*P(B). Thus A and B are not independent.
b) since P(A| C)= 1 and P( B|C)= 1,
P(C)= P(AC)= P(BC)
A | C is a subset of (A U B) | C . Hence P( A U B | C) >= P( A | C) = 1
Hence P( A U B | C) = 1
Hence P( AB | C)= P( A|C) + P( B | C)- P( A U B | C)= 1+1-1= 1
hence P( C) = P( ABC)
let p(C)= p
Now P( A | CC) = P ( ACC)/ P(CC)= (P(A)-p)/(1-p)
and similarly P( B | CC)= (P(B)-p)/(1-p)
and P( AB | CC)= (P(AB)-p)/(1-p)
Now from the assumtion in the question,
(P(AB)-p)/(1-p)= (P(A)-p)/(1-p)*(P(B)-p)/(1-p)
=> 0.1-p = (0.1-p)*(0.25-p)/(1-p) [ p is not equal to 1]
=> (0.1-p)( 1- (0.25-p)/1-p)=0
=> 0.75*(0.1-p)/(1-p)=0
=> p= 0.1
Hence P(C)= 0.1.
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