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Math Help - Probability

  1. #1
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    Probability

    I have two exercises I've got some problems with, and I hope anyone can help me. I know it's possible to use the binominal theorem on the first one, but we haven't learnt that, so I wonder if there is another way of solving it?

    1. A box contains 35 red discs and 5 black discs. A disc is selected at random and its colour noted. The disc is then replaced in the box.

    (a) In eight such selections, what is the probability that a black disc is selected

    (i) exactly once?

    (ii) at least once?


    (b) The process of selecting and replacing is carried out 400 times.
    What is the expected number of black discs that would be drawn?


    2. For the events A and B, p(A) = 0.6, p(B) = 0.8 and p(A U B) = 1

    Find p('A U 'B)
    (complementary)
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  2. #2
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    Hello, MaOp91!

    1. A box contains 35 red discs and 5 black discs.
    A disc is selected at random and its colour noted.
    The disc is then replaced in the box.

    (a) In eight such selections, what is the probability that a black disc is selected:
    . . (i) exactly once?
    . . (ii) at least once?
    Since the discs are replaced: . \begin{array}{ccccc}p(R) &=& \frac{35}{40} &=& \frac{7}{8} \\ \\[-4mm] p(B) &=& \frac{5}{40} &=& \frac{1}{8} \end{array}


    (a) We can solve this without the Binomial Theorem.

    Suppose the outcome is: . BRRRRRRR in that order.
    . . Then the probability is: . \left(\frac{1}{8}\right)\left(\frac{7}{8}\right)^7

    Since the one B can appear in any of eight locations,
    . . p(\text{one B}) \:=\:8 \times \left(\frac{1}{8}\right)\left(\frac{7}{8}\right)^7 \;=\;\frac{6,\!588,\!344}{16,\!777,\!216} \;\approx\;39.3\%



    (b) The opposite of "at least one B" is "no B's" (all R's).

    p(\text{8 R's}) \:=\:\left(\frac{7}{8}\right)^8

    Therefore: . p(\text{at least one B}) \;=\;1 - \left(\frac{7}{8}\right)^8 \;=\;\frac{11,\!012,\!415}{16,\!777,\!216} \;\approx\;65.6\%




    (b) The process of selecting and replacing is carried out 400 times.
    What is the expected number of black discs that would be drawn?

    Since p(B) \,=\,\frac{1}{8}, we would expect get a black disc: . \frac{1}{8} \times 400 \:=\: 50 times.




    2. For the events A and B\!:\;\;p(A) = 0.6,\;p(B) = 0.8,\;p(A \cup B) = 1

    Find: . p(A' \cup B')

    \text{We have:} \;\underbrace{p(A \cup B)}_1 \;=\;\underbrace{p(A)}_{0.6} + \underbrace{p(B)}_{0.8} -\: p(A \cap B) \quad\Rightarrow\quad p(A \cap B) \:=\:0.4


    From DeMorgan's Law: . A' \cup B' \:=\:[A \cap B]'

    Therefore: . p(A' \cup B') \:=\:p\bigg([A \cap B]'\bigg) \:=\:(0.4)' \:=\:0.6

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  3. #3
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    Thanks a loooooot
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