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Math Help - [SOLVED] Probability

  1. #1
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    [SOLVED] Probability

    Seventy percent of kids who visit a doctor have a fever, and 30% of kids with a fever have sore throats. What's the probability that a kid who goes to the doctor has a fever and a sore throat?

    Not sure how to set this one up...

    Thank you for your help!
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  2. #2
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    You set it up exactly like your other question. Do you know the multiplicative properties of probabilities? Your AND or OR statements as it relates to probability?

    First ask yourself, are these events independent or dependent?
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  3. #3
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    These events are dependent?

    So, basing on the AND/OR statements, this is an AND problem, which means
    P(A|B) = P(A) x P(B).

    .70 x .30 = .21 ?
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  4. #4
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    My apologies I read the problem wrong. You got the correct answer however the logic is not right.

    What is being asked here is P(A & B)=P(B|A)P(A).

    Can you figure out what probabilities (.3, .21 and .7) belong to which statements?
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  5. #5
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    I'm sorry...I don't quite understand.

    Do you mean...

    P(.21) = P(.3|.7)P(.7) ?
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  6. #6
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    Ok lets break this down:

    We are told that of a certain space (lets call it sick kids), 0.7 of them will get a fever. We are further told, that OF those gets who get a fever, 0.3 will have a sore throat; in mathematical terms "Given that a kid has a fever, the probability that he will have a fever is 0.3." The question finally asks us, what is the probability that a kid will have a fever AND a sore throat.

    If we break it down even further:

    Suppose I have a box of 10 apples, and I tell you that in that box, 20% will be rotten, how would you go about determining how many rotten apples you have, GIVEN that they came from that box? You would multiply 10 apples by 20% and get 2 apples correct? So what would those 2 apples represent? The two apples would represent apples that came form the box AND are rotten.

    Same applies here: We are given that 0.3 have a sore throat gien they have a fever, and 0.7 have a fever. Plugging that in we do indeed get [P(B|A) = 0.3)][P(A) = 0.7)]=[P(A & B) = 0.21]
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  7. #7
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    Oooohhhh! I see now! Thanks sooo much for that clarification! Probability is tough. I hope I do okay on the test.
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  8. #8
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    No problem. Just remember the basic formula: P(B|A)=P(A & B)/P(A), and be sure to understand just what each of the probabilities are actually referring to. That is the most difficult part of this section of Stats, is understanding what your "Space" is and how to "read" the sentence. Once you get it down it kind of becomes old hat (so long as you don't read it wrong like I did).
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  9. #9
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    Thanks for the advice!

    Just to be clear, the formula for "or" is...

    P(A or B) = P(A) + P(B) - P(A and B)

    right?
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  10. #10
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    That is the correct formula. But again, you want to make sure you understand which probabilities are referring to which portion of that equation. Make sure you understand your sample space. Etc. etc. 95% of these problems are figuring out which numbers refer to which bits, and of that 95% I would say 5% is actually something difficult to suss out.
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  11. #11
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    yeah...that's the hard part. But I'm trying!!
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