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Thread: Conditional Probability

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    Conditional Probability




    I need help with a couple of conditional probability questions:

    If a person is vaccinated properly, the probability of his/her getting a certain disease is 0.05. Without a vaccination, the probability of getting the disease is 0.35. Assume that 1/3 of the population is properly vaccinated. If a person gets the disease, what is the probability that he/she was vaccinated?

    Suppose a test for diagnosing a certain serious disease is successful in detecting the disease in 95% of all person infected, but that it incorrectly diagnoses 4% of all healthy people as having the disease. If it is known that 2% of the population has the disease, find the probability that a person selected at random has the disease if the test indicates that he or she does.

    The probability that a football player weighs more than 230 pounds is 0.69, that he is at least 75 inches tall is 0.55, and that he weighs more than 230 pounds and is at least 75 inches tall is 0.43. Find the probability that he is at least 75 inches tall if he weighs more than 230 pounds.

    I think I must be missing an easy way to figure these out, but I have been scratching my head for a while and could really use a little help.

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    Re: Conditional Probability

    Without looking at the details of the question I guess you may try to apply Bayes theorem first.

    OK - I have to admit that you just need the simple definition of conditional probability for the last one.

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    Re: Conditional Probability

    Well I don't know if this is the right way to do it, but what I tried is this:

    The sample space for the first one is 0.25 (all people with the disease), the outcome of interest is vaccinated people with the disease which is 0.0167 (1/3 * .05), so the probability would be 0.0167/0.25 = 0.0668

    For the second problem, I thought of it like this out of all the diseased people total (0.02), 0.95 of them diagnosed correctly, so the probability would be an "and" situation or 0.95 * 0.02 = 0.019

    And for the last one the sample space would be all fat players (0.69) and the outcome of interest is fat players who are also tall (0.43) so the probability that the player is tall given he is fat is 0.43/0.69 = 0.6232

    But I have to admit I have no idea if I'm doing it correctly.

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    Re: Conditional Probability

    1 and 3 look correct to me. Well, technically I got (1/3 * .05)/.25 =.0\overline{666}

    For 2, I got this:

    \begin{tabular}{|l|c|c|r|}
\hline
 & Diagnosis Yes & Diagnosis No & Total \\
\hline
Disease Yes & .019 & .001 & .02 \\
Disease No  & .0392 & .9408 & .98 \\
\hline
Total & .0582 & .9418 & 1.00 \\
\hline
\end{tabular}

    "Diseased and correctly diagnosed" divided by
    ("Healthy and incorrectly diagnosed" + "Diseased and correctly diagnosed")

    \frac{.95 * .02}{(.04 * .98) + (.95 * .02)} = \frac{.019}{(.0392) + (.019)} = 0.326

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    Re: Conditional Probability


    Thanks! I actually thought more about 2 and came up with the same answer (I made a table and came up with 0.019/0.0582

    Still not positive about the disease/vaccination problem - a fellow student thought it would be 1/3 of all diseased or (1/3*.25) but I think that is more of an and situation and I interpreted the question as a given the person gets the disease type of question (conditional probability) and thought the sample space was now disease people or .25 and the parameter of interest was diseased vaccinated but it's a little confusing.

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