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Math Help - Probability

  1. #1
    Len
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    Probability

    A factory produces items in boxes of 2. Over the long run:
    92% of boxes contain 0 defective items;
    5% of boxes contain 1 defective item;
    3% of boxes contain 2 defectives item.

    A box is picked at random from production, then an item is picked at random from the box. Given that the item is defective, what is the chance the second item in the box is defective.
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  2. #2
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    Quote Originally Posted by Len View Post
    A factory produces items in boxes of 2. Over the long run:
    92% of boxes contain 0 defective items;
    5% of boxes contain 1 defective item;
    3% of boxes contain 2 defectives item.

    A box is picked at random from production, then an item is picked at random from the box. Given that the item is defective, what is the chance the second item in the box is defective.
    Of every 100 boxes, 8 contain at least 1 defective.
    The defective item came from one of these and each box contains only two items.
    There are 6 defectives in the 3 boxes containing 2 defectives each.
    there are 5 defectives in the 5 boxes containing 1 defective each.
    Hence there is a 6 out of 11 chance of having picked a defective item from a box with 2 defectives.
    The next one chosen from that box will be also be defective.
    Therefore, there is a 6 in 11 chance the item was chosen from one of the boxes with 2 defective items.
    Last edited by Archie Meade; January 31st 2010 at 05:17 PM. Reason: error
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  3. #3
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    Hello, Len!

    A factory produces items in boxes of 2.
    Over the long run:
    . . 92% of boxes contain 0 defective items;
    . . 5% of boxes contain 1 defective item;
    . . 3% of boxes contain 2 defectives item.

    A box is picked at random from production, then an item is picked at random from the box.
    Given that the item is defective, what is the chance the second item in the box is defective?
    I changed the language for my convenience . . .


    The factory produces three types of boxes:

    . . Type A: contains no defective items. . P(A) \,=\,0.92

    . . Type B: contains one defective item. . P(B) \,=\,0.05

    . . Type C: contains two defective items. . P(C) \,=\,0.03


    A box is chosen at random and one item is sampled from that box.

    Given that the sample is defective, what the probability that the box is of Type C?


    This is Conditional Probability.

    Bayes' Theorem: . P(\text{Type C }|\text{ 1 d{e}f}) \;=\;  \frac{P(\text{Type C }\wedge\text{ 1 d{e}f})}{P(\text{ 1 d{e}f})} \ .[1]


    Numerator: . P(C) \,=\,0.03

    . . From a Type C box, the probability of drawing a defective item is 100% = 1.

    . . Hence: . P(\text{Type C }\wedge\text{ 1 d{e}f}) \:=\:(0.03)(1) \;=\;0.03 .[2]


    Denominator: . P(B) \,=\,0.05

    . . From a Type B box, the probability of drawing a defective item is 0.5
    . . Hence: . P(\text{Type B }\wedge\text{ 1 d{e}f}) \:=\:(0.05)(0.5) \:=\:0.025

    . . From a Type A box, it is impossible to draw a defective item.

    . . We already know that: . P(\text{Type C }\wedge\text{ 1 d{e}f}) \:=\:0.03

    Hence: . P(\text{ 1 d{e}f}) \;=\; 0.025 +0.03 \:=\: 0.055 .[3]


    Substitute [2] and [3] into [1]: . P(\text{Type  C }|\text{1 d{e}f}) \;=\;\frac{0.03}{0.55} \;=\;\frac{6}{11} \;\approx\;54.5\%

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  4. #4
    Len
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    I do appreciate the explanations, one is intuitive and the other with definition. Thanks so much
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