Math *Questions* Involving Probablity

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Homework Help Overview

The discussion revolves around two probability problems: the first involves determining the number of rolls needed with two six-sided dice to achieve at least a 50% chance of rolling two sixes at least once, and the second concerns calculating the conditional probability of having diabetes given a positive test result from a medical lab.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore different methods for calculating the probability of rolling two sixes, including a reverse approach that focuses on the probability of not rolling two sixes. For the diabetes question, some participants mention using Bayesian probability and others suggest a population-based approach to estimate the conditional probability.

Discussion Status

There are multiple interpretations of the problems being discussed, with participants offering different methods and calculations. Some guidance on using Bayesian equations has been provided, but no consensus has been reached on the final answers.

Contextual Notes

Participants note the lack of formal equations in the original problem statements and express varying levels of familiarity with Bayesian probability concepts.

darshanpatel
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Homework Statement



1) How many times must you roll two six-sided dice for there to be at least a 50% chance that you roll two 6's at least once?2) It is estimated that 5.9% of Americans have diabetes. Suppose a medical lab uses a test for diabetes that 98% accurate for people who have the disease and 95% accurate for the people who do not have it. Find the conditional probability that a randomly selected person actually has diabetes given that the lab test says they have it.

Homework Equations



-None-

The Attempt at a Solution



Question One:

Work: chance of rolling a six(dice one) x chance of rolling a six(dice two) x one half

1/6 x 1/6 x 1/2 = 1/72

72 rolls?

Question Two:

Work:

P(accurate) x (accurate|don't have diabetes) - P(accurate) x (accurate|have diabetes)
(.95 x .941)-(.98 x .059)= .89395-.83163 = about 5.782% ?
 
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q1: mathematicians look at this problem in reverse and instead ask what's the probability of not getting two sixes in a row and then multiply it over and over until they get to the 50% mark and the number of times they multiplied will be the number of rolls needed.

q2: is a a bayesian conditional probability question, you need the bayes eqn to compute it:

p(A given B) = p(B given A) * p(A) / p(B) or succinctly p(A|B)=p(B|A)p(A)/p(B)

where A= prob of having diabetes and B=prob of positive test
 
My approach to problem two was to set an arbitrary number of people in the population (say 1 million) and then figure out the number of people who A) have the disease and get a positive test result and then B) don't have the disease, but still get a positive result.
Then the answer is A/(A+B), which was ~55%I only approached the problem this way because I'm not familiar with bayesian equation that jedishrufu mentioned.
 
Your answer of 5.78% is close to mine of 5.89%
 
darshanpatel said:

Homework Statement



1) How many times must you roll two six-sided dice for there to be at least a 50% chance that you roll two 6's at least once?


2) It is estimated that 5.9% of Americans have diabetes. Suppose a medical lab uses a test for diabetes that 98% accurate for people who have the disease and 95% accurate for the people who do not have it. Find the conditional probability that a randomly selected person actually has diabetes given that the lab test says they have it.


Homework Equations



-None-

The Attempt at a Solution



Question One:

Work: chance of rolling a six(dice one) x chance of rolling a six(dice two) x one half

1/6 x 1/6 x 1/2 = 1/72

72 rolls?

Question Two:

Work:

P(accurate) x (accurate|don't have diabetes) - P(accurate) x (accurate|have diabetes)
(.95 x .941)-(.98 x .059)= .89395-.83163 = about 5.782% ?


Answer for 1 is n = 25. Answer for 2 is 5782/10487 ≈ 0.55, or about 55%.

RGV
 

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