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Various probability problems and their solutions, including the birthday problem, medical diagnosis problem, bertrand's box problem, and let's make a deal problem. These problems cover concepts such as conditional probability and joint probability.
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Four Birthday Problems^2
The general probability is: p = [364/365]n
Since: [364/365]^253 = .499523 ˜. Therefore: 1 - [364/365]^253 = 1-.499523 ˜.
Using: 1-[364/365]n^ = 1/n Ë n ˜ 19
Use: 1-[364/365][363/365][362/365]… [{365-(n-1)}/365] since the nth^ person is the one we’re trying to find the same birthday for.
Let n=23: 1-[364/365][363/365][362/365]… [{365-22}/365] = .492703 ˜.
The Medical Diagnosis Problem^3
D = Has the disease T = Test result is positive
Given: P(D) = .001 so P(Not D) =. P(T | Not D) =. P(T | D) = 1.
(^1) All of these questions are common ones, but I took the suggestion to use them from: Richard G. Seymann. November
Wiley. (^3) This question was presented to 60 students and staff at the Harvard Medical School. Over half of the participants
responded 95% and only 11 gave the correct response. Hmm… See: D. Kahneman, P. Slovik, and A. Tversky. 1982. Judgement Under Uncertainty: Heuristics and Biases. Cambridge, UK: Cambridge University Press.
P(D | T) = P(Tn D) / P(T) P(Tn D) = P(D)P(T | D) =. P(T) = P(Tn D) + P(Tn Not D) = P(Tn D) + P(Not D)P(T | Not D) = .001 + (.999)(.05) =. P(D | T) = .001/.05095 =.
“Bertrand’s Box” Problem^4
Initial set-up
P(A | G) = P(An G)/P(G) = (1/3)/(1/2) = 2/
(^4) See: R. Weatherford. 1982. Philosophical Foundations of Probability Theory. London: Routledge and Kegan Paul.
Now, the host shows you what’s behind a door (assume that it’s door #2) and asks, “Do you want to switch doors (from door #1 to door #3)?”
P(Car Behind Door #1 | Host Shows Door #2) = P(Car D1 | Shows D2) = P(Car D1 n Shows D2)/P(Shows D2) = (1/6)/(1/2) = 1/ P(Car Behind Door #3 | Host Shows Door #2) = P(Car D3 | Shows D2) = P(Car D3 n Shows D2)/P(Shows D2) = (1/3)/(1/2) = 2/
Since P(Car Behind Door #1 | Host Shows Door #2) < P(Car Behind Door #3 | Host Shows Door #2), you should switch from door #1 to door #3.