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Math Help - probability....

  1. #1
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    probability....

    Hello,
    please try to solve this question.Thanks.
    Question no:1
    The probability that a patient recovers from a rare blood disease is 0.4. If 15 people are known to have contracted this disease, what is the probability that:

    (a)at least 10 survive
    (b)from 3 to 8 survive
    (c)exactly 5 survive
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by m777 View Post
    Hello,
    please try to solve this question.Thanks.
    Question no:1
    The probability that a patient recovers from a rare blood disease is 0.4. If 15 people are known to have contracted this disease, what is the probability that:

    (a)at least 10 survive
    (b)from 3 to 8 survive
    (c)exactly 5 survive
    The number of survivors has a binomial distribution B(15, 0.4), so:

    <br />
p(n\ survivors) = b(n;15,0.4)=\frac{15!}{n!(15-n)!}(0.4)^n(1-0.4)^{15-n}<br />

    (a) p(10\ or\ more\ survivors) = \sum_{n=10}^{15}b(n;15,0.4)

    (b) p(3\ to\ 8\ survivors) = \sum_{n=3}^{8}b(n;15,0.4)

    (c) p(5\ survivors) = b(5;15,0.4)

    RonL
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  3. #3
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    Hello, m777!

    Who asks these questions?
    Do they really expect us to crank out these answers?


    The probability that a patient recovers from a rare blood disease is 0.4.
    If 15 people are known to have contracted this disease, what is the probability that:

    (a) at least 10 survive
    (b) from 3 to 8 survive
    (c) exactly 5 survive

    (a) "At least 10" means: 10 or 11 or 12 or 13 or 14 or 15.

    P)10) \:=\: {15\choose 10}(0.4)^{10}(0.6)^5
    P(11) \:=\: {15\choose11}(0.4)^{11}(0.6)^4
    P(12) \:=\: {15\choose12}(0.4)^{12}(0.6)^3
    P(13) \:=\: {15\choose13}(0.4)^{13}(0.6)^2
    P(14) \:=\: {15\choose14}(0.4)^{14}(0.6)^1
    P(15) \:=\: {15\choose15}(0.4)^{15}(0.6)^0

    . . Now add them up . . .



    (b) "From 3 to 8" means: 3 or 4 or 5 or 6 or 7 or 8.

    P(3) \:=\:{15\choose3}(0.4)^3(0.6)^{12}
    P(4) \:=\:{15\choose4}(0.4)^4(0.6)^{11}
    P(5) \:=\:{15\choose5}(0.4)^5(0.6)^{10}
    P(6) \:=\:{15\choose6}(0.4)^6(0.6)^9
    . \vdots\qquad\qquad\qquad\vdots

    . . You get the idea . . .



    (c) Exactly 5 . . . We did this one in part (b).

    P(5)\:=\:\frac{15!}{5!10!}(0.4)^5(0.6)^{10} \:=\:0.0612979282

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, m777!

    Who asks these questions?
    Do they really expect us to crank out these answers?



    (a) "At least 10" means: 10 or 11 or 12 or 13 or 14 or 15.

    P)10) \:=\: {15\choose 10}(0.4)^{10}(0.6)^5
    P(11) \:=\: {15\choose11}(0.4)^{11}(0.6)^4
    P(12) \:=\: {15\choose12}(0.4)^{12}(0.6)^3
    P(13) \:=\: {15\choose13}(0.4)^{13}(0.6)^2
    P(14) \:=\: {15\choose14}(0.4)^{14}(0.6)^1
    P(15) \:=\: {15\choose15}(0.4)^{15}(0.6)^0

    . . Now add them up . . .



    (b) "From 3 to 8" means: 3 or 4 or 5 or 6 or 7 or 8.

    P(3) \:=\:{15\choose3}(0.4)^3(0.6)^{12}
    P(4) \:=\:{15\choose4}(0.4)^4(0.6)^{11}
    P(5) \:=\:{15\choose5}(0.4)^5(0.6)^{10}
    P(6) \:=\:{15\choose6}(0.4)^6(0.6)^9
    . \vdots\qquad\qquad\qquad\vdots

    . . You get the idea . . .



    (c) Exactly 5 . . . We did this one in part (b).

    P(5)\:=\:\frac{15!}{5!10!}(0.4)^5(0.6)^{10} \:=\:0.0612979282

    My answer is 0.015603175...i using this formula
    10C5 x (0.4)^5 x (0.6)^10...
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  5. #5
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    None of them will survive. In 200 years, they will all die...I'm 99% sure of that.

    Edit: I've checked and double checked my data and it appears I had made a mistake... it should be 87.3%.
    Last edited by rualin; July 9th 2007 at 10:09 AM.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by rualin View Post
    None of them will survive. In 200 years, they will all die...I'm 99% sure of that.
    That 99% should by tradition be 87.3% (since as all professionals know 87.3% of all statistics are made up)

    RonL
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