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Math Help - Some Probability Questions - Needing Help

  1. #1
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    Question Some Probability Questions - Needing Help

    Hello, I have these probability problems to solve for my math homework, but am having some problems with the methodology. The answers are provided, but I need a detailed step-by-step process to see the way to arrive at the solution. Any help is greatly appreciated!

    Here are the questions:

    1) There are 50 units of some product, with 5 defective units. 3 units are randomly chosen, what is the probability of choosing exactly 1 defective unit?

    2) From the clubs of a deck of cards (13 cards), randomly drawing 3 card while returning the card after each draw, what's the probability of picking a card of a different value each and every time?

    3) There are 2 white balls and 3 black balls in a bag. Withdrawing two balls, one ball at a time, what's the probability of both balls being white?

    4) Two machines are processing the same parts, the machine one has a defect rate of 0.03, machine two has a defect rate of 0.02. The processed parts are grouped together, and machine one processed twice as many parts as machine two. What's the probability of randomly choosing a non-defective product? If the chosen product is defective, what's the probability that it was processed by machine two?

    5) There are 12 ping pong balls in a box, 9 of the 12 are new. For the first match, 3 are randomly chosen, then returned to the box. For the second match, 3 are again randomly chosen. What's the probability of choosing all new balls for the second match? If the balls chosen for the second match were new, what's the probability of having chosen all new balls for the first match?

    Thanks a lot!
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  2. #2
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    1. The first one is given by a hyper geometric distribution.

     \frac{45c2 \cdot 5c1}{50c3} = \frac{99}{392}

    2. I think you are saying just a 13 card deck of cards. So after drawing the first card you have a \frac{12}{13} chance of not drawing the same card. On the third draw you have a \frac{11}{13} chance of not drawing either of the previous 2 cards. The product of the two probabilities is your answer, that is: \frac{132}{169}


    3. P(first ball being white)  \cdot P(2nd ball being white) = \frac{2}{5} \cdot \frac{1}{4} = \frac{1}{10} (tree diagram)

    4. P(defective) =  \frac{0.02 + 2(0.03)}{2} = 0.04. So the probability its not defective is the complement of this:  .96 If the chosen product is defective, what's the probability that it was processed by machine two?(conditional probability) P(From machine 2| Defective)  \frac{0.02}{0.04} = \frac{1}{2}

    5. P(choosing all new balls for the 2nd match) = P(B) = \frac{9}{12} \cdot  \frac{8}{11} \cdot  \frac{7}{19}=  \frac{21}{55}
    Let P(choosing all new balls for the 1st match) = P(A). We want the P(A|B) = P(AB)/P(B). But because these 2 events are independent P(A|B) = P(A) =  \frac{21}{55}

    If someone wants to check if im right that would be nice
    Last edited by Scopur; December 6th 2008 at 09:53 PM.
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