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Math Help - Simple Probability Problem

  1. #1
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    Exclamation Simple Probability Problem

    In a particular town 60% of the population are women. 4% of the men and 1% of the women are taller than 180 cm.

    i) What % of the town's population is taller than 180 cm?

    ii) If a person is chosen at random and is taller than 180 cm, what is the probability that the person is a woman?

    Ummm I'm not that good at probability so could someone please give me some help?

    For:

    i) I did:

    Percentage = 100(0.04*0.4 + 0.01*0.6)
    = 0.022*100
    = 2.2%

    ii) I did:

    P = 0.022*0.2
    = 11/2500

    The answers don't look rite though...
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  2. #2
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    You have the first part right, for the second one you need to use Baye's theorem.


    <br />
P(woman|taller than 180)= \frac{P(taller than 180|woman)P(woman)}{P(taller than 180|woman)P(woman)+P(taller than 180|man)P(man)}<br />
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  3. #3
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    Quote Originally Posted by xwrathbringerx View Post
    In a particular town 60% of the population are women. 4% of the men and 1% of the women are taller than 180 cm.

    i) What % of the town's population is taller than 180 cm?

    ii) If a person is chosen at random and is taller than 180 cm, what is the probability that the person is a woman?

    Ummm I'm not that good at probability so could someone please give me some help?

    For:

    i) I did:

    Percentage = 100(0.04*0.4 + 0.01*0.6)
    = 0.022*100
    = 2.2% Mr F says: Correct.

    ii) I did:

    P = 0.022*0.2 Mr F says: Wrong. Where has the 0.2 come from?
    = 11/2500

    The answers don't look rite though...
    ii requires conditional probability.

    Here is the magic recipe, the big secret, the great trick, the golden elixir. Read it carefully ........

    Set up careful notation and definitions. Carefully define all known probabilities in terms of this notation.

    Let T be the event height greater than 180 cm.

    You have:

    Pr(M) = 2/5, Pr(W) = 3/5,

    Pr(T | M) = 1/25, Pr(T | W) = 1/100,

    Pr(M and T) = Pr(T| M) Pr(M) = (1/25) (2/5) = 2/125,

    Pr(W and T) = Pr(T| W) Pr(W) = (1/100) (3/5) = 3/500,

    Pr(T) = Pr(M and T) + Pr(W and T) = 2/125 + 3/500 = 11/500 (= 0.022 as you found).

    You require Pr(W | T).


    \Pr(W | T) = \frac{\Pr(W \cap T)}{\Pr(T)} = \frac{3/500}{11/500} = 3/11.
    Last edited by mr fantastic; October 10th 2008 at 10:20 PM. Reason: Added a line
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  4. #4
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    You have the first part right, for the second one you need to use Baye's theorem.
    To keep not that messy, Let
    W=woman
    M=Man
    T=taller than 180

    <br />
P(W|T)= \frac{P(T|W)P(W)}{P(T|W)P(W)+P(T|M)P(M)}=\frac{(0.  01)(0.60)}{(0.01)(0.60)+(0.04)(0.40)}<br />

    <br />
=\frac{0.006}{0.006+0.016}=\frac{0.006}{0.022}=\fr  ac{3}{11}<br />
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