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Thread: True/False Probability Problems

  1. #1
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    True/False Probability Problems




    Can you confirm that these answers are correct? Thank you.
    6. True
    7. False
    8. False
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  2. #2
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    = 1
    2/3 + 2/3 - 4/9 =
    4/3 - 4/9 =
    12/9 - 4/9 =
    9/9 = 1
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  3. #3
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    Hello, krzyrice!

    Here are the first two . . . I'm working on the third one.


    Let $\displaystyle A$ and $\displaystyle B$ be two events in a finite sample space.
    True or false?

    6. If $\displaystyle P(A) + P(B) \:=\:1$, then $\displaystyle A$ and $\displaystyle B$ must be complementary. . True

    We have: .$\displaystyle P(B) \:=\:1 - P(A)$

    Hence: .$\displaystyle B \:=\:A'$ . . . They are complementary.



    7. For all $\displaystyle A$ and $\displaystyle B, \;P(A \cap B) + P(A \cup B) \:=\:P(A) + P(B)$ . True
    This is a just rearrangement of the well-known formula:

    . . . . $\displaystyle P(A \cup B) \:=\:P(A) + P(B) - P(A \cap B) $

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  4. #4
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    Sorry, Plato, I must disagree . . .


    The problem gave: .$\displaystyle {\color{blue}P(A) + P(B) \:=\:1}$

    There was no mention of an "or", as in $\displaystyle A \cup B$


    You have: .$\displaystyle \begin{array}{ccccccc}P(A) &=& P(1,2,3,4) &=& \frac{2}{3} \\ \\[-4mm]P(B) &=& P(3,4,5,6) &=& \frac{2}{3} \end{array}$

    . . . . and: .$\displaystyle P(A) + P(B) \:\neq \:1$

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  5. #5
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    Okay, Plato, you win!

    I had to come up with a truly embarrassing example to convince me.


    Event A: draw card from a deck and get a Heart.

    .$\displaystyle P(A) \:=\:\frac{13}{52} \:=\:\frac{1}{4}$


    Event B: flip two coins and get at least one Head.

    . . $\displaystyle P(B) \:=\:\frac{3}{4}$


    Hence, $\displaystyle P(A) + P(B) \:=\:1$ . . . but $\displaystyle A$ and $\displaystyle B$ are not complementary.


    Well, duh!

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  6. #6
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    thanks for the help guys .
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