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Thread: I am stuck in this probability question and i need some help to find out

  1. #1
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    Unhappy I am stuck in this probability question and i need some help to find out

    Spares of a particular component are produced by two firms , Bestbits and Lesserprod. Tests
    show that, on average, I in 200 components produced by Bestbits fail within one year Of
    fitting, and I in 50 components produced by Lesserprod fail within One year Of fitting.
    Given that 20 per cent of the components sold and fitted are made by Bestbits and 80 per
    cent by Lesserprod, what is the probability that a component chosen at random from those
    sold and fitted will fail within a year of fitting?
    Find the proportion of components sold and fitted that would need to be made by Bestbits
    for this probability to be 0.01 .
    (OCR)
    Attached Thumbnails Attached Thumbnails I am stuck in this probability question and i need some help to find out-capture.png  
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  2. #2
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    Re: I am stuck in this probability question and i need some help to find out

    $P[fail|B]=0.005$

    $P[fail|L]= 0.02$

    $P[fail] = P[fail|B]P[B] + P[fail|L]P[L] = (0.005)(0.2)+(0.02)(0.8) = 0.017$

    $(0.005)P[B] + (0.02)(1-P[B]) = 0.01$

    I leave you to solve for $P[B]$
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    Re: I am stuck in this probability question and i need some help to find out

    thank You <3
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  4. #4
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    Re: I am stuck in this probability question and i need some help to find out

    Imagine this happening 1000 times.
    "Given that 20 percent of the components sold and fitted are made by Bestbits and 80 percent by Lesserprod".
    So 200 are from "Bestbits" and 800 are from "Lesserprod".

    "I in 200 components produced by Bestbits fail within one year Of
    fitting"
    (You mean "1 in components", not "I in 200"!) So of the 200 from best bits, one fails.

    "and I in 50 components produced by Lesserprod fail within One year Of fitting."
    So of the 800 from Lesserprod, 800/50= 16 fail.

    That is a total of 17 out of 1000 failing or 17/1000= 0.017 or 1.7%.


    More generally, assume proportion "p" were from Bestbits and 1- p were from LesserProd. Then of the 1000, 1000p are from Bestbits and 1000(1- p) are from Lesserprod. Again, "1 in 200" from Bestbits, 1000p/200= 5p fail. And "1 in 50" from Lesserprod 1000(1- p)/50= 200(1- p) fail. That is a total or 5p+ 200- 200p= 200- 195p of the 1000 or (200- 195p)/1000. We want that to be 0.01. So solve (200- 195p)/1000= 0.01.
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  5. #5
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    Re: I am stuck in this probability question and i need some help to find out

    Quote Originally Posted by Deadpool View Post
    Spares of a particular component are produced by two firms , Bestbits and Lesserprod.

    Tests show that, on average, 1 in 200 components produced by Bestbits fail within one year
    of fitting, and 1 in 50 components produced by Lesserprod fail within one year of fitting.

    Given that 20 per cent of the components sold and fitted are made by Bestbits and 80 per
    cent by Lesserprod, what is the probability that a component chosen at random from those
    sold and fitted will fail within a year of fitting?

    Find the proportion of components sold and fitted that would need to be made by Bestbits
    for this probability to be 0.01
    .....making it EASIER to read!!
    Last edited by DenisB; Apr 15th 2019 at 09:13 AM.
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    Re: I am stuck in this probability question and i need some help to find out

    Thanks Buddy for helping out
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