Results 1 to 4 of 4

Math Help - Probability problem

  1. #1
    Newbie
    Joined
    Jun 2006
    Posts
    4

    Probability problem

    1> A company manufactures a product which consists of 2 parts, part A and part B. If 8 out of 100 are likely to be defective in part A, similarly 6 out of 100 are to be defective for part B. Calculate the probability that the assumed part will not be defective?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by electronic_pen
    1> A company manufactures a product which consists of 2 parts, part A and part B. If 8 out of 100 are likely to be defective in part A, similarly 6 out of 100 are to be defective for part B. Calculate the probability that the assumed part will not be defective?
    The probability that Part A will not be defective is \frac{100}{100}-\frac{8}{100}=\frac{92}{100}

    The probability that part B will not be defective is \frac{100}{100}-\frac{6}{100}=\frac{94}{100}

    Now you multiply those together to get the probability that both parts will not be defective
    \frac{92}{100}\times\frac{94}{100}

    =\frac{92\times 94}{100\times 100}

    =\frac{8,648}{10,000}

    =\frac{86.48}{100}

    so the answer is \frac{86.48}{100} or 86.48%
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,914
    Thanks
    779
    Hello, electronic_pen!

    The wording is strange . . . I'm not sure I understand the question
    . . but I'll make a guess.


    1) A company manufactures a product which consists of 2 parts: part A and part B.
    8 out of 100 are likely to be defective in part A.
    Similarly, 6 out of 100 are to be defective for part B.
    calculate the probability that the assumed part ? will not be defective.

    Since P(\text{A d{e}f}) = \frac{8}{100}, then P(\text{A not d{e}f}) = \frac{92}{100}

    Since P(\text{B d{e}f}) = \frac{6}{100}, then P(\text{B not d{e}f}) = \frac{94}{100}


    If a randomly chosen product is not defective,
    . . then part A is not defective and part B is not defective.

    P(\text{A not d{e}f} \land \text{B not d{e}f}) \;= \;\frac{92}{100}\cdot\frac{94}{100}\;=\;\frac{1081  }{1250}


    Edit: You're simply too quick, Quick!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jun 2006
    Posts
    4

    Thanks: QuIcK and sorobon, btw value for QuIcK is as follows,

    The problem can be perceived by in both the ways. First one, finding the probability of those assumed(here the defective proportion are assumed) parts(A and B) are not defective. [and] second one, the probability that those assumed parts WILL not be defective , i mean the probability of the respective defective parts not equalling their assumption(here 8/100 and 6/100).

    For the second possibility, i think we should make use of hypothesis and compute standard error(sqrt(pq/n)), here (sqrt(.08*.92/100)) and finding z (normal probability distribution). If Z ie .(assumed proportion-experimenting proportion)/standard error) is > or = 1.96 then the assumption is rejected and viceversa. Atleast one which proves the defective proportion is more or less than the assumed one, we can conclude those assumed ones are wrong, and their probability is a no of favourable trial to no of possibilities (100). For the experimenting proportion, it is just the ramdom number from 1 to 100 other than 8 in the first case.

    However, i got headache in proving by the second method.
    Guys thanks for the problem, solved in a simple way.

    For the QuIcK value,

    the value gets reduced to 512000*c/(L*o)

    if c = correctness in percentage
    L = reciprocal of Labourious in percentage
    o = objectivity in percentage.

    QuIcK = 512000*(.95)/(1.25*.95)
    =409000
    =409K

    Proficiency = 409/512 * 100
    = 80%

    Thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: January 21st 2011, 12:47 PM
  2. Probability Problem
    Posted in the Statistics Forum
    Replies: 3
    Last Post: January 7th 2010, 11:49 PM
  3. Replies: 0
    Last Post: October 8th 2009, 09:45 AM
  4. Probability problem 4
    Posted in the Statistics Forum
    Replies: 1
    Last Post: March 10th 2009, 12:38 AM
  5. A Probability Problem
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: December 27th 2008, 06:17 PM

Search Tags


/mathhelpforum @mathhelpforum