Tuesday, May 22, 2007

Exactly why are doctors so cagey?!

Hey it's MrSiwWy here to present to you the scribe post for today! I'm the scribe for today's rather intriguing but yet solemn class. I say solemn as I think that the main question our class was focused around pertained to very serious matter. But anyways, onto the scribe post for
Tuesday, May 22, 2007.

Our class began with Mr. K outright stating to the class that we're on the verge of the end of probability, and the beginning of our final unit: sequences. He then told us that he won't be present the two following days, but that we only have two things left to learn in probability, one of which incorporates one of our earlier units: counting and combinatorics. He presented us with an example of such a question. This question was regarding the arrangement of students, like we have been presented with in the past.
If students are to be seated in 7 desks, what is the probability that Mark and Robert are sitting together?
Basically, we take amount of favorable outcomes, which would be the amount of ways Robert and Mark can be seated together, then divide that by the total amount of seating possibilities.

We quickly moved onto the class for today, and very soon, we transitioned into the question that Mr. K left us off with on Thursday; "Why do doctors have do so many tests? Why are they so cagey?" Once Mr. K put up the question we were to dissect, he set us up in small groups of four. This smoothly yet appropriately forwarded the momentum of our class into the actual lesson. This question can be seen on the following slide:

If we think about each of the four possibilities, we have to realize that the test is to check for cancer, so if it tests positive when the person really does have cancer, then the test succeeded, meaning that it has a probability of 98%, or 0.98. This means the remaining probability (that the test failed by showing negative, when they really have cancer) is 2%, or 0.02.
But as for people who do not have cancer, if the test shows positive, this means that the test failed, and this probability is in fact 2%, or 0.02, and the remaining probability (to test negative for those who do not have cancer) is 98%, or 0.98.
And as I stated at the end of the slide, a tree diagram can really help. So here is a tree diagram that was used to calculate the probabilities of each of the questions on the following slide.

Mr. K later asked us to construct this tree diagram to help us understand the question's calculations, but really, to create one beforehand is more of a help.

Here we see that Mr. K coded certain important aspects for various reasons. These reasons are very vital when contemplating a method to approach each question.
The blue underlined portions indicate that we're dealing with people who have cancer.
The green underlined portions indicate that we're dealing with people who do not have cancer.
The small circular flower symbols indicate total amounts of people that have cancer.
The small stars indicate total amounts of people that have been tested positive for cancer.
The last two are particularly important because the main question that we were originally asked was what the probability of a patient that has cancer is also tested positive for cancer. In order to arrive at this, we must dignify how many total people have cancer, and divide that by the amount of people that were tested positive. We've already learned that in order to find the probability of an event, we take the amount of favorable outcomes and divide that by the total amount of outcomes. So in this case, we take the amount of subjects who have cancer, and divide that by the total amount of times that the test would show a positive result. That is why Mr. K signified each of the above portions. This was taken care of in part 4 of the above questions, but before that, I'd like to outline how to achieve each of the answers in the entire slide.
In order to determine each of the answers shown on this slide, we used the tree diagram above in conjunction with the sample group of 1,000,000 people.

  1. This part was simple. Since we know that there are 1,000,000 people, and 0.5% of these people have cancer, that means that 5000 people have cancer, and the rest don't. The rest can be found by subtracting how many that have cancer from the total of people, this gives us 995000.
  2. In this part we use the tree diagram to determine what percent of people that have cancer will test positive. To do this we multiplied the amount of people that have cancer, then multiplied that by the probability of testing positive, which is 0.98. This gave us 4900, the answer presented in the slide. For b, since we know 4900 tested positive, the rest of the people who have cancer must have tested negative, and that number is 100.
  3. Here we repeated the same process used in part 2, exact that we used different values since we're looking at only the people who do not have cancer. We know that there are 995000 people that do not have cancer, and as I outlined earlier, since the test would fail if it said positive, we would have to find 2% of 995000. This, as shown, is 19900, and to find how many tested negative, we just subtract this from the total. 995000-19900 gives us an answer of 975100.
  4. The final part was compounding what we determined in the previous steps, and in essence, determining the answer to our original question. First we take how many people altogether tested positive, we get this simply by adding those that have cancer that tested positive together with those that don't have cancer but tested positive. For b, we know that of the people that tested positive, 4900 of them actually had cancer. So to find the original question at hand, this was the work shown. I already described how this was found, here is the work:
As he showed us that vertical line separation between the C and the P, the class began wondering what that represented. This introduced the class to conditional probability. The way Mr. K described such a probability was "when we have inside information" beforehand. Usually when we have dealt with probability we knew that either one thing or another can happen, then another event can happen in one way or another. But in conditional probability, we know that one of these events is going to happen in such a way. In this question, we knew that our answer only concerns when we know that the test is going to be positive. He gave another example of conditional probability using three jars containing marbles. He says that if jar one has 3 blue and 3 red marbles, jar 2 has 2 blue and 3 red, jar 3 has 2 blue and 2 red, then he doesn't look at any of the jars and pulls out a blue marble out of one of them. In this case, we know that the marble that he pulled out is blue. Given the above information, what is the probability that it is from jar 2 of pulling a blue marble from jar 2? This isn't word for word, but I believe that was the basis of the example Mr. K conveyed to us. The basic idea is that we know that the marble is going to be blue, similar to how we knew that the test was to be positive in the first question today. So to sum up conditional probability:
Conditional probability is when we have some inside information on the events at hand, basically when we know some of the information before we even start to calculate.
When we are dealing with conditional probability we write P(P1P2), where P1 is the probability that we're looking for, the vertical line says "given that we already know", and P2 is the event that we already know.

Once we were finished with that question, we moved onto a very similar question, in which he just changed "cancer" to "industrial condition," 98 % to 99 % and 0.5% to 1%. Here is the question along with the work, of which I will of course explain.

Here we see both a tree diagram and the work to the solution for the question.
  • I represents those with industrial disease.

  • H represents those without industrial disease.

  • P represents those that tested positive for the disease.

  • N represents those that tested negative for the disease.
Here we see that we found the probability that someone has the disease and also tests positive for the disease by dividing the favourable outcome (probability of having the disease) by the total amount of testing positive. This is exactly the same case as finding the probability of having cancer and testing positive for cancer, just with different probabilities.

So in this class, we fully determined the answer to the question that Mr. K had left us with on Thursday afternoon.

Doctors aren't necessarily cagey, but run the tests more than once to make sure. If we look at the results that we achieved by both questions, we arrive at the fact that if someone really has the disease and tests positive, there's not a good chance that it's for sure. Thus, doctors run tests more than once not for fun or anything of the sort, but in both cases it's more likely or equally likely to be tested positive and not have the disease, so it's good to be sure before breaking such news to a person.

And that concluded our class for today, with the lesson of conditional probability that was presented to us today, as Mr. K said, we don't have a lot left in this unit. This was my scribe post, and I hope it was clear enough for anyone to understand. I still need to fix some content up, really don't like how my Computer is so resilient, especially seeing as I can't get to any pod casts. =/ But as soon as my computer begins to cooperate with me I'll get everything fixed up and tidy. Please just contact me if there are any concerns, questions, complaints, errors, etc. Have a good night everybody!
oh yes, thank you John for reminding me. The scribe for tomorrow will be: Jann!


John D. said...

Err.. MrSiWwy, who's the next scribe? haha

MrSiwWy said...

That's right, almost forgot to pick one. But it's so hard to choose, it's not up to date and I don't know who has scribed third cycle already and who hasn't =/ I just picked Jann because he sits behind me =) I love my reasoning.