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Math Help - Probability

  1. #1
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    Probability

    A manufacturer of teddy bears has three assembly lines: 1, 2 and 3. The percentages of total daily output that are produced by the lines are 25%, 35% and 40%, respectively. The percentages of defective units produced by the lines are estimated to be 1%, 2% and 1% respectively. If a teddy bear is randomly selected from a day's production, what is the probability that it is defective?
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  2. #2
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    Quote Originally Posted by MathGeek06 View Post
    A manufacturer of teddy bears has three assembly lines: 1, 2 and 3. The percentages of total daily output that are produced by the lines are 25%, 35% and 40%, respectively. The percentages of defective units produced by the lines are estimated to be 1%, 2% and 1% respectively. If a teddy bear is randomly selected from a day's production, what is the probability that it is defective?
    i could be wrong...but i think the answer is

    .25(.01) + .35(.02) + .40(.01)
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    Another Probability Question

    On Friday a store advertised a special sale for Saturday. On Saturday it was found that 60% of customers had known about the sale and, of these customers, 40% bought a sale item. Of the customers that had not known about the sale, 20% bought a sale item. If a customer bought a sale item, what is the probability that he or she had known about the sale?
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  4. #4
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    using similar logic...
    .6(.4) + .4(.2) will get you the % of everyone who bought a sale item.

    now just divide that number by the percentage of people who knew about the sale.
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  5. #5
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    Quote Originally Posted by MathGeek06 View Post
    On Friday a store advertised a special sale for Saturday. On Saturday it was found that 60% of customers had known about the sale and, of these customers, 40% bought a sale item. Of the customers that had not known about the sale, 20% bought a sale item. If a customer bought a sale item, what is the probability that he or she had known about the sale?
    Let K be the event that the customer knows about the sale.
    Let B be the event that the customer buys an item.

    P(K | B) = \frac{P(B|K)P(K)}{P(B|K)P(K) + P(B|K^c)P(K^c)}

    P(K | B) = \frac{(0.4)(0.6)}{(0.4)(0.6) + (0.2)(0.4)}

    P(K | B) = 0.75
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  6. #6
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Dubulus View Post
    i could be wrong...but i think the answer is

    .25(.01) + .35(.02) + .40(.01)
    This one seems correct at a glance, but....

    Quote Originally Posted by Dubulus View Post
    using similar logic...
    .6(.4) + .4(.2) will get you the % of everyone who bought a sale item.

    now just divide that number by the percentage of people who knew about the sale.
    ...this one does not look correct to me.
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  7. #7
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by Dubulus View Post
    using similar logic...
    .6(.4) + .4(.2) will get you the % of everyone who bought a sale item.

    now just divide that number by the percentage of people who knew about the sale.
    Janvdl is right, this is incorrect (not by much though)

    to find this percentage, you need to divide the number of people who knew about the sale and bought a sale item by the number of people who bought a sale item.

    .6(.4) = .24 so 24% of people bought a sale item and knew about the sale

    .4(.2) = .08 so 8% of people bought a sale item but did not know about the sale

    So then your ratio is: \frac{.24}{.24+.08}=\frac{.24}{.32}=\frac{3}{4}=.7  5=\boxed{75\%}
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  8. #8
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    Thank u

    THANKS SO MUCH EVERYONE!!!!!!
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