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Probability of selecting defective light bulbs

  1. Jan 29, 2009 #1
    1. The problem statement, all variables and given/known data

    A box contains 24 light bulbs of which 4 are defective. If one person selects 10 bulbs from the box in a random manner, and a second person then takes the remaining 14 bulbs, what is the probability that all 4 detective bulbs will be obtained by the same person.?


    2. Relevant equations

    I don't know how to show the solution because I can't find the big parentheses and numbers are one above the other (24 10) 10 from 24 Combinatorial Methods

    3. The attempt at a solution

    Here we 24 light bulbs in that 4 aare defectives.

    First person takes 10 bulbs from the box that can be done in the (10 form 24) ways.

    In that 4 are defective items remeaning are not defective items.

    probability that all 4 detective bulbs will be obtained by the same person
    Required probability
     
    Last edited: Jan 29, 2009
  2. jcsd
  3. Jan 29, 2009 #2
    Re: probability

    probability (4 from 4)*(6 from 20)/(10 from 24) + (10 from 20)/(10 from 24).

    Does someone know how to write this in form for combinatorial methods. I want to verify my answer.

    Please help.
     
  4. Jan 29, 2009 #3
    Re: probability

    Looks correct to me. To calculate (24 10) for example; use the [tex]_n C_r[/tex] on your calculator: [tex]24 \ _nC_r \ 10[/tex].

    Probability that the first person selects all the four defective ones:

    (4 4)(20 6)/(24 10) = 0.01976

    Probability that the second person selects all the four defective ones:

    (4 0)(20 10)/(24 10) = 0.09420

    Then you can add the two probabilities because they are independent occurences:

    P = 0.1140
     
    Last edited: Jan 29, 2009
  5. Jan 29, 2009 #4
    Re: probability

    Thank you. I forgot to check the the answer in the textbook, it is given only :

    (20 6) + (20 10) / (24 10).

    Thank you again.
     
  6. Jan 29, 2009 #5
    Re: probability

    where is the remaining 14. In that 4 are defective items remaining are not defective items.
    then:

    (4 4)(20 6) / (24 10) + (20 10) /(24 14)=0.114

    that way answer is the same but not the same in the book.
     
  7. Jan 29, 2009 #6
    Re: probability

    Just two different ways of regarding the same problem. The probability of choosing 14 (including all the 4 deficient ones) out of 24, is the same as choosing 10 normal bulbs out of 24.
     
  8. Jan 29, 2009 #7
    Re: probability

    Thanks.
     
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