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Math Help - Probability

  1. #1
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    Probability

    Hi, I was wondering if anyone could help me out with this question:
    I really have no idea where to proceed.

    Billy enters a shed, and close the door. Inside the shed there are 6 mosquitoes.
    One of them carries the Ross River virus.
    a) Billy is bitten by a mosquito which he slaps and kills. He is then bitter by another mosquito, and again slaps and kills the offender. Find the probability that Billy was bitten by the mosquito with the virus.

    b) (A new situation, still with 6 mosquitoes).
    While in the shed, Billy was bitten twice. Calculate the probability that Billy was bitten by the virus carrying mosquito at least once, given that he swiped but missed after feeling each bite.

    Any help is much appreciated!
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  2. #2
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    Quote Originally Posted by Gaarleaf View Post
    Hi, I was wondering if anyone could help me out with this question:
    I really have no idea where to proceed.

    Billy enters a shed, and close the door. Inside the shed there are 6 mosquitoes.
    One of them carries the Ross River virus.
    a) Billy is bitten by a mosquito which he slaps and kills. He is then bitter by another mosquito, and again slaps and kills the offender. Find the probability that Billy was bitten by the mosquito with the virus.

    b) (A new situation, still with 6 mosquitoes).
    While in the shed, Billy was bitten twice. Calculate the probability that Billy was bitten by the virus carrying mosquito at least once, given that he swiped but missed after feeling each bite.

    Any help is much appreciated!
    a) Pr(RR') + Pr(R'R) = \frac{1}{6} + \frac{5}{6} \times \frac{1}{5} = \frac{1}{3}.

    b) 1 - Pr(not bitten by virus carrying mosquito).
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  3. #3
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    You've made alot clearer, thanks.

    But am I right in thinking that Part B can be done using binomial distribution, ie X~B(2, 1/6), P(x >/= 1) = 11/36?

    Or, the other way would be to consider the 3 cases;

    (1/6 x 5/6) + (5/6 x 1/6) + (1/6 x 1/6)?
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  4. #4
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    Quote Originally Posted by Gaarleaf View Post
    You've made alot clearer, thanks.

    But am I right in thinking that Part B can be done using binomial distribution, ie X~B(2, 1/6), P(x >/= 1) = 11/36?

    Or, the other way would be to consider the 3 cases;

    (1/6 x 5/6) + (5/6 x 1/6) + (1/6 x 1/6)?
    Yes. But using 1 - Pr(X = 0) is clearly the simplest approach.
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