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Probability that bulb is defective questions

  1. Jan 23, 2013 #1
    1. The problem statement, all variables and given/known data
    There are 100 light bulbs in a box. 20 are 40 watt, the rest are 60 watt.
    If we randomly choose 10 bulbs, what is the probability that there will be exactly 2 40-watt bulbs?

    What is the probability that in a random sample of 10 bulbs there will be at least 1 40-watt bulb?

    and then i am also having trouble with this one


    75% of a certain part is supplied by vendor A and 25% by vendor B. Vendor A's defect rate is 0.01 and vendor B's defect rate is 0.03.

    If a part is from vendor A, what is the probability it is defective?

    What is the probability that a randomly chosen part will be defective and from vendor A?

    What is the probability that a randomly chosen part will be non-defective and from Vendor B?

    3. The attempt at a solution

    I think that these are nCr problems but I am very confused by these problems, if someone could explain them to me that would be amazing, thank you so much.
     
  2. jcsd
  3. Jan 23, 2013 #2

    HallsofIvy

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    The probability that the first is 40 watt is, of course, 20/100= .2. That would leave 99 bulbs with 19 of them 40 watt. The probability that the second bulb is also 40 watt is 19/99 so that both first and second will be 40 watt is (20/100)(19/99). That leaves 98 bulbs of which 80 are NOT 40 watt so the probability the third is NOT 40 watt is 80/98, the probabiility the fourth is not 40 watt is 79/97, the probability the fifth is not 40 watt is 78/96, etc. That is the probability of the first two being 40 Watt and the other 8 not is (20/100)(19/99)(80/98)(79/97)...(73/91).

    If you apply the same analysis to, say, "4th and 6th bulbs 40 watt all others not" or "3rd and 9th bulbs 40 watt others not", etc. you should soon realize that you are just moving numbers in the numerator around and, in fact, get the same number repeatedly so you really just have to multiply that first number by the number of ways to put 2 40 watt bulbs and 8 non- 40 watt bulbs which is the "binomial coefficent" 2C10.

    This is actually easier. "At least one 40-wtt bulb" is the opposite of "all 60 watt" and so is 1 minus the probability that all 10 are 60 watt.

    This doesn't require any calculation- it is just given: "Vendor A's defect rte is 0.01"

    I would imagine 10000 parts. We are told that 75% or 7500 are from vendor A and 1% of those, or 75 are defective. So the probability is 75/1000= .075.

    Same idea. If there were 1000 parts, 25% or 250 are from vendor B. 3% of those, 7.5 are defective so the probability is 7.5/1000.

     
    Last edited: Jan 23, 2013
  4. Jan 23, 2013 #3
    Thank you so much, that was very helpful and cleared up some things.
     
  5. Jan 23, 2013 #4
    My only question is for the first one you said to multiply .2* 2C10, however when i try to do 2C10 i get 0, or an error
     
  6. Jan 23, 2013 #5

    haruspex

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    Halls meant 10C2, the number of ways of choosing 2 items from 10.
    ... and he didn't say the answer was .2 * 10C2.
    In terms of combinations, you have to choose 2 from the 10 and 8 from the 80. How many ways does that make?
    That is compared with all the possible choices, namely, 10 from 100.
     
    Last edited: Jan 23, 2013
  7. Jan 23, 2013 #6
    for 10C2 i'm geting 45, if i multiply 45*.2 I get 9, doesnt my answer have to be between 0 and 1?
     
  8. Jan 23, 2013 #7

    haruspex

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    My edit wasn't quite fast enough. See what I added in the previous post.
     
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