Alg1231W10T4 - Tree diagrams and medical test

Friday, 28 September 2018

3:01 PM

If a medical test for a condition returns 'positive', it means @ according to the test, the person has the condition O that the situation is good - the patient is okay. Medical tests are O always accurate @ not always accurate. The success rate of a medical test O is independent of whether the patient actually has the condition @ may vary depending on whether the patient does or does not have the condition.

 

A certain medical test to detect a particular disease (D) has a 99% chance of returning positive (Y) if a person has the disease, and a 98% chance of returning negative (N) if a person does not have the disease. Suppose 3% of the population has the disease. This situation is described by the following tree diagram: D 0.99 0.03 0.97 0.01 0.02 DC 0.98 Recall that P(YID) is a conditional probability, and denotes @ the probability of getting a positive result (Y) on the test, given that the person has the disease (D) U the probability of having the disease (D) given that the person gets a positive result (Y) on the test. From either the diagram or the numbers provided, we see that 0.03 O P(DC) 0.97 P(NID) - P(YIDC) = P(NIDC) -

 

0.03 0.97 i) The probability of having the disease and testing positive is n Y) = P(D) P(YID) = ii) The probability of not having the disease and testing positive is P(DC n Y) = P(DC) P(YIDC) = 0.99 0.01 0.02 iii) The total probability of testing positive is, by the Total Probability Rule, the sum of these, namely P(Y) — P(D) P(YID) + P(DC) P(YIDC) = The conditional probability Of D given Y is, by Bayes• Rule, P(DIY) = so there is roughly a 40% 0.6048879837 O chance that you do not have the disease even if you do test positive for it. This is an important and perhaps surprising result that is worth understanding: just because you test positive for something does not guarantee that you actually have it. Note: your answers should be entered with at_least 2 significant figures Of accuracy.

 

Let's continue with the previous scenario: 0.03 0.97 Calculate P(DCIN) = 0.99 0.01 0.02 Note: your answers should be entered with at_least 2 significant figures of accuracy.

 

There are three different types of circus prizes marked big (B), medium (M) and little (L). Each contains a certain number of red (R) and gold (C) balls, distributed as follows • big prize (B) : 3R and 5G • medium prize (M) : '2R and 2G • little prize (L) : IR and IC Your friend wins 3 big prizes, 1 medium prize and 2 little prizes. Without looking, you randomly reach into one Of her prizes, and randomly take out one Of its balls, which happens to be gold (G). Calculate the probability that you were choosing from a little prize bag. P(LIG) = Note: your answer should be entered with at-least 2 significant figures. Hint: formulate the question as a conditional probability.

 

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