HSC 2015 MX1 Marathon (archive) (1 Viewer)

[InteGrand]

Re: HSC 2015 3U Marathon

 

[leehuan]

Re: HSC 2015 3U Marathon

Overlooked the turning point at x=1 despite having the graph open. Tsk, I will fix that up.

 

[davidgoes4wce]

Re: HSC 2015 3U Marathon

I have a question its a probability based question.

A medical report shows that the carriers of a particular disease during winter months constitute 2% of the population.

Lisa has developed a blood test which is 90% accurate in detecting both presence and absence of the disease in a person. Show that the probability of a false positive result (i.e that a non-carrier is incorrectly identified as being a carrier) is 0.098. Given a person gets a positive result, what is the probability that t he person is a non-carrier?

 

[InteGrand]

Re: HSC 2015 3U Marathon

I have a question its a probability based question.

A medical report shows that the carriers of a particular disease during winter months constitute 2% of the population.

Lisa has developed a blood test which is 90% accurate in detecting both presence and absence of the disease in a person. Show that the probability of a false positive result (i.e that a non-carrier is incorrectly identified as being a carrier) is 0.098. Given a person gets a positive result, what is the probability that t he person is a non-carrier?

FIRST PART:

Show that the probability of a false positive result (i.e that a non-carrier is incorrectly identified as being a carrier) is 0.098.

Pr(a randomly selected result is false positive) = Pr(The person tested was a non-carrier)×Pr(The test failed (which is 10% regardless of whether the person is a carrier or not)) = 0.98 × 0.1 = 0.098.

SECOND PART:

Given a person gets a positive result, what is the probability that t he person is a non-carrier?

Bayes' theorem tells us that

.

For the purposes of our question,

A = the event that a person is a non-carrier
B = the event that a person gets a positive result.

Then we have:

Pr(B|A) = Pr(positive result | person is non-carrier) = 0.098 (from the first part),
Pr(A) = Pr(person is non-carrier) = 0.98,
Pr(B) = Pr(person gets a positive result) = Pr(person is a carrier)×Pr(positive result | person is carrier) + Pr(non-carrier)×Pr(positive result | non-carrier) = 0.02×0.9 + 0.98×0.1 = 0.116.

Plugging these in, our answer is .

So if you get a positive result, there's actually about an 82.8% chance you don't have the disease!

 

[InteGrand]

Re: HSC 2015 3U Marathon

(This calculation and result is assuming the disease is independently and identically distributed throughout the population.)

 

[davidgoes4wce]

Re: HSC 2015 3U Marathon

Thanks InteGrand really appreciate your reply.

I personally did have a go at it as well and came up with this:




I drew a tree diagram to work out the individual probabilities

 

[davidgoes4wce]

Re: HSC 2015 3U Marathon

FIRST PART:



Pr(a randomly selected result is false positive) = Pr(The person tested was a non-carrier)×Pr(The test failed (which is 10% regardless of whether the person is a carrier or not)) = 0.98 × 0.1 = 0.098.

SECOND PART:


Bayes' theorem tells us that

.

For the purposes of our question,

A = the event that a person is a non-carrier
B = the event that a person gets a positive result.

Then we have:

Pr(B|A) = Pr(positive result | person is non-carrier) = 0.098 (from the first part),
Pr(A) = Pr(person is non-carrier) = 0.98,
Pr(B) = Pr(person gets a positive result) = Pr(person is a carrier)×Pr(positive result | person is carrier) + Pr(non-carrier)×Pr(positive result | non-carrier) = 0.02×0.9 + 0.98×0.1 = 0.116.

Plugging these in, our answer is .

So if you get a positive result, there's actually about an 82.8% chance you don't have the disease!

My thinking was the 90% positive result implied that it could be for it being the combination of both carrier and non-carrier combined. In your calculation you have used 0.90 for it being a positive result given its a carrier

 

[swagswagyoloswag]

Re: HSC 2015 3U Marathon

FIRST PART:



Pr(a randomly selected result is false positive) = Pr(The person tested was a non-carrier)×Pr(The test failed (which is 10% regardless of whether the person is a carrier or not)) = 0.98 × 0.1 = 0.098.

SECOND PART:


Bayes' theorem tells us that

.

For the purposes of our question,

A = the event that a person is a non-carrier
B = the event that a person gets a positive result.

Then we have:

Pr(B|A) = Pr(positive result | person is non-carrier) = 0.098 (from the first part),
Pr(A) = Pr(person is non-carrier) = 0.98,
Pr(B) = Pr(person gets a positive result) = Pr(person is a carrier)×Pr(positive result | person is carrier) + Pr(non-carrier)×Pr(positive result | non-carrier) = 0.02×0.9 + 0.98×0.1 = 0.116.

Plugging these in, our answer is .

So if you get a positive result, there's actually about an 82.8% chance you don't have the disease!

The results for probability questions always go against my intuition. that's such a large % of not carrying it when the prob. was 0.098 of being false positive

 

[davidgoes4wce]

Re: HSC 2015 3U Marathon

This may be another stupid question but why cant we use the conditional probability instead?

P(A|B)=P(A AND B)/ P(B) instead? everything kind of makes sense until that last step

 

[InteGrand]

Re: HSC 2015 3U Marathon

My thinking was the 90% positive result implied that it could be for it being the combination of both carrier and non-carrier combined. In your calculation you have used 0.90 for it being a positive result given its a carrier

Usually the percentage they give is the probability the test will be correct, i.e. 0.9 chance that it gives positive given the person is a carrier, and 0.9 chance it gives negative given the person is a non-carrier.

 

[davidgoes4wce]

Re: HSC 2015 3U Marathon

Correct me if I'm wrong (I think I am probably wrong)






I drew a tree diagram




Start 0.02 for Carrier (0.90 for Positive and 0.10 for Negative)
OR
0.98 for Non-Carrier (0.10 for Positive and 0.90 for Negative)




P(Positive Result)= 0.90 x 0.02 + 0.98 x 0.10=0.116 (same value as you Integrand)

I am reading the question as:

P(Non-Carrier|Positive Result)=P(Non Carrier AND Positive Result)/P(Positive Result)= [0.10 x 0.98 ]/ 0.116 = 0.844 or 49/58

 

[InteGrand]

Re: HSC 2015 3U Marathon

This website may help:
http://vassarstats.net/bayes.html

(By the way, I think conditional probability is not in the HSC syllabus.)

 

[davidgoes4wce]

Re: HSC 2015 3U Marathon

Thanks for your help Integrand. These questions(positive/negative) always sting me but your explanation cleared a few things up . Cheers!

 

[InteGrand]

Re: HSC 2015 3U Marathon

Also this: http://betterexplained.com/articles/an-intuitive-and-short-explanation-of-bayes-theorem/

 

[InteGrand]

Re: HSC 2015 3U Marathon

The results for probability questions always go against my intuition. that's such a large % of not carrying it when the prob. was 0.098 of being false positive

If you haven't already, you should check out the Monty Hall Problem (famous probability problem)!

 

[VBN2470]

Re: HSC 2015 3U Marathon

Another fascinating problem in Probability is known as the the Birthday Problem (or Birthday Paradox). It may seem rather counter-intuitive at first, but it essentially computes the probability of a pair of persons sharing the same birthday given a set of people in a room. The mind-blowing thing about it is that there only needs to exist a group of 70 people in a room to obtain a 99.9% chance of any pair sharing the same birthday, whilst you must have a group of 366 (or 367) people for there to exists a 100% chance of any pair sharing the same birthday. Interesting stuff! In some way, this shows that humans as a species tend to have a relatively poor probabilistic sense, in that we can't necessarily make fairly accurate predictions about the possibilities of events that can occur, possibly due to the effect of past experiences which cloud our ability to make these correct judgements. So, the mathematics we devise to represent and work with these probabilities helps us compute these chances in a more well-defined manner.

 

[InteGrand]

Re: HSC 2015 3U Marathon

Another fascinating problem in Probability is known as the the Birthday Problem (or Birthday Paradox). It may seem rather counter-intuitive at first, but it essentially computes the probability of a pair of persons sharing the same birthday given a set of people in a room. The mind-blowing thing about it is that there only needs to exist a group of 70 people in a room to obtain a 99.9% chance of any pair sharing the same birthday, whilst you must have a group of 366 (or 367) people for there to exists a 100% chance of any pair sharing the same birthday. Interesting stuff! In some way, this shows that humans as a species tend to have a relatively poor probabilistic sense, in that we can't necessarily make fairly accurate predictions about the possibilities of events that can occur, possibly due to the effect of past experiences which cloud our ability to make these correct judgements. So, the mathematics we devise to represent and work with these probabilities helps us compute these chances in a more well-defined manner.

Marilyn vos Savant (famed for holding the Guinness World Record for highest IQ, before that record was discontinued) got a probability Q wrong on her regular column.

And new question:

 

[Drsoccerball]

Re: HSC 2015 3U Marathon

Marilyn vos Savant (famed for holding the Guinness World Record for highest IQ, before that record was discontinued) got a probability Q wrong on her regular column.

And new question:

 

[davidgoes4wce]

Re: HSC 2015 3U Marathon

I checked with another tutor today Integrand, and he reckons you did everything right except for the value of 0.098

P(A|B) = (0.10 x 0.98 )/0.116

 

[Drsoccerball]

Re: HSC 2015 3U Marathon

But that violates Bayes' Theorem? Here's a specific example related to medical testing: http://mechanism.ucsd.edu/teaching/philpsych/inevitableillusion3.pdf

But I think Bayes' Theorem and conditional probability are not in the course.

Bae's theorem ?