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Math Help - Please help me with this challenging math question probabilty

  1. #1
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    Question Please help me with this challenging math question probabilty

    Please answer any of the following question and please show your work

    For [ 2x^2 - 1/x]^6, determine
    a) The middle term
    b) The term containing x^3

    Given the universal set S = {-3,-2,-1,0,1,2,3,4} and the sets A = {1,2,3,4}, B = {-3,-2,-1}, find C = {-2,-1,0,1,2}
    FInd
    a) A'
    b) (A n C) and (B n C)
    c) (B n C)'

    Your are dealt one card from a standard deck of playing cards.
    a) What is the probability that the card is 5?
    b) What is the probability that the card is a 8 or a club?

    Eight percent of the objects produced by a machine are defective. If you examine 50 objects produced by the machine, find the probability that
    a) 3 are defective
    b) More than 3 are defective

    A class consists of 8 boys and 5 girls
    a) In how many ways can a teacher choose a committee of 3 from this class?
    b) How many of these committees will contain at least 1 boy?
    c) How many will contain exactly 1 girl
    What is the probability that number chosen from 1 to 500 is divisible by 2, 5 or 6?

    A box contains 5 black balls, 4 red balls, 4 green balls, and 5 blue balls.
    a) YOu select 6 balls from the box without replacing them. What is the probabilty that
    you will select excatly 3 black balls?
    b) You select 6 balls from the box without replacing them. What is the probability that
    you draw at least 1 black balls?
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  2. #2
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    Hello, john002!

    For \left(2x^2 - \tfrac{1}{x}\right)^6, determine

    (a) the middle term

    (b) the term containing x^3
    We are expected to know the Binomial Theorem.

    \left(2x^2 - \tfrac{1}{x}\right)^6 \;=\;{6\choose6}(2x^2)^6\left(\text{-}\tfrac{1}{x}\right)^0 + {6\choose5}(2x^2)^5\left(\text{-}\tfrac{1}{x}\right)^1 + {6\choose4}(2x^2)^4\left(\text{-}\tfrac{1}{x}\right)^2 + {6\choose3}(2x^2)^3\left(\text{-}\tfrac{1}{x}\right)^3 + {6\choose2}(2x^2)^2\left(\text{-}\tfrac{1}{x}\right)^4 + {6\choose1}(2x^2)^1\left(\text{-}\tfrac{1}{x}\right)^5 + {6\choose0}(2x^2)^0\left(\text{-}\tfrac{1}{x}\right)^6

    . . . . . . . =\; (1)(64x^{12})(1) \;+\; 6(32x^{10})\left(\text{-}\tfrac{1}{x}\right)\;+\;15(16x^8)\left(\tfrac{1}{  x^2}\right) \;+ \;20(8x^6)\left(\text{-}\tfrac{1}{x^3}\right) \;+\; 15(4x^4)\left(\tfrac{1}{x^4}\right) . +\;6(2x^2)\left(\text{-}\tfrac{1}{x^5}\right) + 1(1)\left(\tfrac{1}{x^6}\right)

    . . . . . . . =\;64x^{12} - 192x^9 + 240x^6 \underbrace{-\; 160x^3}_{(a),\;(b)} + 60 - \frac{12}{x^3} + \frac{1}{x^6}




    You are dealt one card from a standard deck of playing cards.

    a) What is the probability that the card is 5?

    b) What is the probability that the card is a 8 or a club?
    There are 52 cards in a standard deck of cards.


    (a) Four of them are 5's.

    . . P(5) \:=\:\frac{4}{52} \:=\:\frac{1}{13}


    (b) There are four 8's: . 8\heartsuit,\;8\spadesuit,\;8\diamondsuit,\;8\club  suit
    . . .There are 13 Clubs: . A\clubsuit,\;2\clubsuit,\;3\clubsuit,\;4\clubsuit,  \;5\clubsuit,\;6\clubsuit,\;7\clubsuit,\;8\clubsui  t,\;9\clubsuit,\;10\clubsuit,\;J\clubsuit,\;Q\club  suit.\;K\clubsuit

    . . .But don't count the 8\clubsuit twice!

    Hence, there are 16 cards that are an 8 or a Club.

    . . P(\text{8 or Club}) \:=\:\frac{16}{52} \:=\:\frac{4}{13}




    A box contains 5 black balls, 4 red balls, 4 green balls, and 5 blue balls.
    You select 6 balls from the box without replacing them.

    (a) What is the probabilty that you select excatly 3 black balls?
    There are: . {18\choose6} \,=\,18,\!564 possible outcomes.

    We have: .5 Black balls and 13 Others
    We want: 3 Black balls and 3 Others.

    There are: . {5\choose3} = 20 ways to pick 3 Black.

    There are: . {13\choose5} = 286 ways to pick 3 Others.

    Hence, there are: . 20 \times 286 \:=\:5,\!720 ways to pick 3 Blacks and 3 Others.

    Therefore: . P(\text{at least 3 Blacks}) \:=\:\frac{5,\!720}{18,\!564} \;=\;\frac{110}{357}



    (b) What is the probability that you draw at least 1 black ball?
    The opposite of "at least one Black" is "no Blacks".

    To get no Blacks, we want 6 of the 13 Others: . {13\choose6} = 1716 ways.
    . . Hence: . P(\text{no Blacks}) \:=\:\frac{1716}{18,\!564} \:=\:\frac{11}{119}


    Therefore: . P(\text{at least one Black}) \;=\;1 - \frac{11}{119} \;=\;\frac{108}{119}

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