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Math Help - Probability

  1. #1
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    Probability

    A four digit number is to be made by choosing four different digits from the set {1,2,3,4,5,6,7}. Find the probability that the no. is even and greater than 6000.
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  2. #2
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    Hello, BabyMilo

    A 4-digit number is to be made by choosing four different digits from the set {1,2,3,4,5,6,7}.
    Find the probability that the number is even and greater than 6000.
    There are: . _7P_4 \:=\:\frac{7!}{3!} \:=\:840 possible 4-digit numbers.

    How many of them are even and greater than 6000?

    There are two cases to consider.


    [1] The number begins with "6": .6 _ _ _

    We have: .{1, 2, 3, 4, 5, 7} to work with.

    The last digit must be even: 2 choices.

    The middle two digits are selected from the remaining 5 digits: . _5P_2= 20 choices.

    Hence, there are: . 2\cdot20 = 40 even numbers that begin with "6".


    [2] The number begins with "7": .7 _ _ _

    We have: .{1, 2, 3, 4, 5, 6} to work with.

    The last digit must be even: 3 choices.

    The middle two digits are selected from the remaining 5 digits: . _5P_2 = 20 choices.

    Hence, there are: . 3\cdot20 = 60 even numbers that begin with "7".


    Then there are: . 40 + 60 \:=\:100 even numbers greater than 6000.


    Therefore: . P\bigg[(\text{even}) \wedge (> 6000)\bigg] \;=\;\frac{100}{840} \:=\:\frac{5}{42}

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