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Math Help - 2 Questions----I need clarification.

  1. #1
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    2 Questions----I need clarification.

    The following are 2 practice questions for my math class. There will be similar questions on my test so I need to know how to do them correctly. Any help would be great, thanks!

    1) Following are 3 boxes containing letters: [tex] [AND] [HISTORY].

    a) From box 1, three letters are drawn one-by-one without replacement and recorded in that order. What is the probability that the outcome is HAT?

    -I did (1/4)(1/3)(1/2) = 1/24 Is this correct?

    b) If a box is chosen at random and then a letter is drawn at random from the box, what is the probability that the outcome is A?

    -I'm thinking along the lines of (2/3)(1/4)(1/3), but am unsure...


    2) Box 1: [2 black balls & 3 white balls] Box 2: [4 black balls & 2 white balls] A ball is drawn at random from box 1, then at random from box 2. The colors of the balls are recorded in order. The boxes look like this:
    1:[BBWWW] 2:[BBBBWW] Find:

    a) The probability of 2 whites.

    -I did (3/5)(2/6) = 1/5 Is this correct?

    b) The probability of at least one black. (I'm stumped)

    c) The probability of at most one black. (I'm stumped again)

    d) The probability of a black followed by a white, or a white followed by a black.

    -I did (2/5)(2/6) = 2/15 or (3/5)(4/6) = 2/5 Is this correct?
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  2. #2
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    Hello, mdenham2!

    1) There are 3 boxes containing letters: .  [M,A,T,H]\;\; [A,N,D]\;\; [H,I,S,T,O,R,Y].

    a) From box 1, three letters are drawn one-by-one without replacement.
    What is the probability that the outcome is HAT?

    I did: . \frac{1}{4}\cdot\frac{1}{3}\cdot\frac{1}{2} \,=\,\frac{1}{24} . . . Is this correct? . . Yes!

    b) If a box is chosen at random and then a letter is drawn at random from the box,
    what is the probability that the outcome is A?
    The box is chosen at random: . P(\text{box 1}) \,=\,P(\text{box 2}) \,=\,P(\text{box 3}) \,=\,\frac{1}{3}

    From box 1, P(A) = \frac{1}{4}
    . . Hence: . P(\text{box 1 }\wedge \text{ A}) \:=\:\frac{1}{3}\cdot\frac{1}{4} \:=\:\frac{1}{12}

    From box 2, P(A) = \frac{1}{3}
    .Hence: . P(\text{box 2 }\wedge\text{ A}) \:=\:\frac{1}{3}\cdot\frac{1}{3} \:=\:\frac{1}{9}

    From box 3, P(A) = 0
    . . Hence: . P(\text{box 3 }\wedge\text{ A}) \:=\:\frac{1}{3}\cdot0 \:=\:0


    Therefore: . P(A) \;=\;\frac{1}{12} + \frac{1}{9} + 0 \;=\;\frac{7}{36}




    2) Box 1: [2 black balls & 3 white balls]
    . . Box 2: [4 black balls & 2 white balls]

    A ball is drawn at random from box 1, then at random from box 2.

    a) The probability of 2 whites.

    I did: . \frac{3}{5}\cdot\frac{2}{6} \,=\, \frac{1}{5} . . . Is this correct? . Yes!

    b) The probability of at least one black.
    The opposite of "at least one black" is "no blacks" (both white).

    You found: P(\text{2 whites}) = \frac{1}{5} .in part (a).

    Therefore: . P(\text{at least one black}) \;=\;1-\frac{1}{5} \;=\;\frac{4}{5}



    c) The probability of at most one black.
    The opposite of "at most one black" is "both black".

    P(\text{both black}) \:=\:\frac{2}{5}\cdot\frac{4}{6} \:=\:\frac{4}{15}

    Therefore: . P(\text{at most one black}) \;=\;1-\frac{4}{15} \;=\;\frac{11}{15}




    d) The probability of a black followed by a white, or a white followed by a black.

    I did : . \frac{2}{5}\cdot\frac{2}{6} \,=\,\frac{2}{15}\:\text{ or }\:\frac{3}{5}\cdot\frac{4}{6} \,=\,\frac{2}{5} . . . Is this correct? . Um, not quite ...
    They are expecting one answer.

    With an "or" statement, we add the probabilties.

    Therefore: . P(\text{BW }\vee\text{ WB}) \;=\;\frac{2}{15} + \frac{2}{5} \;=\;\frac{8}{15}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Here's an alternate approach to part (d).


    We want: . P(\text{opposite colors})


    In part (a), we found that: . P(\text{2 whites}) \,=\,\frac{1}{5}

    In part (c), we found that: . P(\text{2 blacks}) \,=\,\frac{4}{15}

    . . Hence: . P(\text{same color}) \:=\:\frac{1}{5} + \frac{4}{15} \:=\:\frac{7}{15}


    Therefore: . P(\text{opposite colors}) \;=\;1 - \frac{7}{15} \;=\;\frac{8}{15}

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  3. #3
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    Thank you!

    Thank you so much, I couldn't have asked for better explanations!
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  4. #4
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    Thanks!!!
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