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Probability Problems and Solutions, Assignments of Statistics

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.

Typology: Assignments

Pre 2010

Uploaded on 08/09/2009

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Interesting Probability Questions1
Four Birthday Problems2
1. Assume that there are n people in the room. Ignoring leap years, what is the probability that no one
else in the room shares your birthday?
The general probability is: p = [364/365]n
2. Assume that there are 253 people in the room. Ignoring leap years, what is the probability that no
one else in the room shares your birthday? What is the probability that someone else in the room
shares your birthday?
Since: [364/365]253 = .499523 ˜ .5
Therefore: 1 - [364/365]253 = 1-.499523 ˜ .5
3. Assume that there are n people in the room. Ignoring leap years, what value of n (most closely)
makes the probability that someone else shares your birthday (1/n)?
Using: 1-[364/365]n = 1/n è n ˜ 19
4. Assume that there are n people in the room. Ignoring leap years, what value of n (most closely)
makes the probability that two people in the room share birthdays equal to .5?
Use: 1-[364/365][363/365][362/365][{365-(n-1)}/365]
since the nth person is the one were trying to find the same birthday for.
Let n=23: 1-[364/365][363/365][362/365][{365-22}/365] = .492703 ˜ .5
The Medical Diagnosis Problem3
5. Assume that a test to detect a disease whose prevalence is (1/1000) has a false positive rate of 5%
and a true positive rate of 100%. What is the probability that a person found to have a positive result
actually has the disease assuming that you know nothing about the persons symptoms?
D = Has the disease
T = Test result is positive
Given:
P(D) = .001 so P(Not D) = .999
P(T | Not D) = .05
P(T | D) = 1.00
1All of these questions are common ones, but I took the suggestion to use them from: Richard G. Seymann. November
1991. Comment”. The American Statistician. 45(4):287-8.
2 Due to: William Feller. 1957. An Introduction to Probability Theory and its Applications (Volume 1). New York: John
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. HmmSee: D. Kahneman, P. Slovik, and A. Tversky. 1982.
Judgement Under Uncertainty: Heuristics and Biases. Cambridge, UK: Cambridge University Press.
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Interesting Probability Questions^1

Four Birthday Problems^2

  1. Assume that there are n people in the room. Ignoring leap years, what is the probability that no one else in the room shares your birthday?

The general probability is: p = [364/365]n

  1. Assume that there are 253 people in the room. Ignoring leap years, what is the probability that no one else in the room shares your birthday? What is the probability that someone else in the room shares your birthday?

Since: [364/365]^253 = .499523 ˜. Therefore: 1 - [364/365]^253 = 1-.499523 ˜.

  1. Assume that there are n people in the room. Ignoring leap years, what value of n (most closely) makes the probability that someone else shares your birthday (1/n)?

Using: 1-[364/365]n^ = 1/n Ë n ˜ 19

  1. Assume that there are n people in the room. Ignoring leap years, what value of n (most closely) makes the probability that two people in the room share birthdays equal to .5?

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

  1. Assume that a test to detect a disease whose prevalence is (1/1000) has a false positive rate of 5% and a true positive rate of 100%. What is the probability that a person found to have a positive result actually has the disease assuming that you know nothing about the person’s symptoms?

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

  1. “Comment”. The American Statistician. 45 (4):287-8. (^2) Due to: William Feller. 1957. An Introduction to Probability Theory and its Applications (Volume 1). New York: John

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

  1. A box has three drawers; one contains two gold coins, one contains two silver coins, and one contains one gold coin and one silver coin. Assume that one drawer is selected randomly and that a randomly selected coin from that drawer turns out to be gold. What is the probability that the chosen drawer is one that contains two gold coins?

Initial set-up

  1. Assuming that the boxes are selected randomly: A = Box A (2 gold coins) P(A) = P(Pick Box A) = 1/ B = Box B (1 gold & 1 silver coin) P(B) = P(Pick Box B) = 1/ C = Box C (2 silver coins) P(C) = P(Pick Box C) = 1/
  2. Since there are six coins, 3 of each type, and any one coin can be selected randomly: P(G) = P(Pick a Gold Coin) = P(Pick a Silver Coin) = P(Not G) = ½
  3. Given (1) and (2), the marginal probabilities on the outside of the joint probability table must be the following: Pick a ____ Coin Pick Box ___ Gold Silver P(Pick Box) Box A P(A)=1/ Box B P(B)=1/ Box C P(C)=1/ P(Pick Coin) P(G)=1/2 P(Not G)=1/2 1
  4. Further, given the description of each box in (1) the joint probability table must be the following: Pick a ____ Coin Pick Box ___ Gold Silver P(Pick Box) Box A 1/3 0 P(A)=1/ Box B 1/6 1/6 P(B)=1/ Box C 0 1/3 P(C)=1/ P(Pick Coin) P(G)=1/2 P(Not G)=1/2 1

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.