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Math Help - probability 3

  1. #1
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    probability 3

    Of the coffee makers sold in an appliance store, 6.0% have either a faulty switch or a defective cord, 2.4% have a faulty switch, and 0.3% have both defects. What percent of the coffee makers will have a defective cord?
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  2. #2
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    Hello, arslan!

    We are expected to know the formula:

    . . P(A \cup B) \;=\; P(A) + P(B) - P(A \cap B)



    Of the coffee makers sold in an appliance store,
    6.0% have either a faulty switch or a defective cord,
    2.4% have a faulty switch, and 0.3% have both defects.
    What percent of the coffee makers will have a defective cord?
    Let S = faulty switch, C = faulty cord.

    We are told that: . \begin{array}{ccc}P(S \cup C) &=& 0.06 \\ P(S) &=& 0.024 \\ P(S \cap C) &=& 0.03 \end{array}


    We have: . P(S \cup C) \;=\;P(S) + P(C) - P(S \cap C)

    . . . . . . . . . . 0.06 \;\;\;= \;\;0.024 + P(C) \;\;-\;\; 0.03


    Therefore: . P(C) \:=\:0.066

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, arslan!

    We are expected to know the formula:

    . . P(A \cup B) \;=\; P(A) + P(B) - P(A \cap B)


    Let S = faulty switch, C = faulty cord.

    We are told that: . \begin{array}{ccc}P(S \cup C) &=& 0.06 \\ P(S) &=& 0.024 \\ P(S \cap C) &=& 0.03 \end{array}


    We have: . P(S \cup C) \;=\;P(S) + P(C) - P(S \cap C)

    . . . . . . . . . . 0.06 \;\;\;= \;\;0.024 + P(C) \;\;-\;\; 0.03


    Therefore: . P(C) \:=\:0.066
    my answers are multiple choice and this answer is not one of the options?
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  4. #4
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    Quote Originally Posted by arslan View Post
    my answers are multiple choice and this answer is not one of the options?
    Soroban made a simple careless mistake in converting one of the percentages into a decimal but his logic is perfectly sound. Your job is to find that simple error, fix it and then get the correct answer.
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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, arslan!

    We are expected to know the formula:

    . . P(A \cup B) \;=\; P(A) + P(B) - P(A \cap B)


    Let S = faulty switch, C = faulty cord.

    We are told that: . \begin{array}{ccc}P(S \cup C) &=& 0.06 \\ P(S) &=& 0.024 \\ P(S \cap C) &=& 0.03 \end{array}


    We have: . P(S \cup C) \;=\;P(S) + P(C) - P(S \cap C)

    . . . . . . . . . . 0.06 \;\;\;= \;\;0.024 + P(C) \;\;-\;\; 0.03


    Therefore: . P(C) \:=\:0.066
    Quote Originally Posted by mr fantastic View Post
    Soroban made a simple careless mistake in converting one of the percentages into a decimal but his logic is perfectly sound. Your job is to find that simple error, fix it and then get the correct answer.
    got it 3.9
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  6. #6
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    Quote Originally Posted by arslan View Post
    got it 3.9
    3.9% is correct.
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