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Math Help - [SOLVED] Set probability questions

  1. #1
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    [SOLVED] Set probability questions

    1 ] The events A and B are such that
    P(A) = 0.2, P(B) = 0.6 and P(A|B) = 0.3.
    Find
    (a) P(A∩B), [2]
    (b) P(B|A), [3]
    (c) P(A∪B). [3]
    (d) P(A‘∩B‘). [2]

    2] The events A and B are such that P(A) = 0.5 and P(A∪B) = 0.7. Determine the value of P(B) in each of the cases when
    (a) A and B are mutually exclusive, [2]
    (b) A and B are independent, [4]
    (c) P(B|Α) = 0.3. [3]

    My answers are:

    1]
    a] 0.18
    b] 0.9
    c] 0.62
    d] 0.82

    2]
    a] 0.2
    b] Cannot understand how to work out

    Sorry I couldn't post working, I have it all written down and it was difficult enough to write all of the questions down! If I am wrong, could you please give me a clue towards getting it right and could you please explain how to attempt 2b]?

    Thanks if you can answer!
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  2. #2
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    Hello, db5vry!

    As we say in The Corps, ya done good!


    The events A and B are such that

    P(A) = 0.2,\;\;P(B) = 0.6,\;\;P(A|B) = 0.3

    Find:

    (a)\;P(A\cap B)\quad{\color{blue}0.18}\quad{\color{red} \hdots \text{ Yes!}}

    (b)\;P(B|A)\quad{\color{blue}0.9}\quad{\color{red} \hdots \text{ Right!}}

    (c)\;P(A \cup B)\quad{\color{blue}0.62}\quad {\color{red} \hdots \text{ Correct!}}

    (d)\:P(A' \cap B')\quad{\color{blue}0.82}\quad{\color{red}\hdots \text{ no}}

    A' \cap B' is the complement of  A\cup B

    So: . P(A' \cap B') \:=\:1 - 0.62 \:=\:\boxed{0.38}




    2] The events A and B are such that: P(A) = 0.5,\;P(A \cup B) = 0.7

    Determine the value of P(B) in each of the cases when:

    (a) A and B are mutually exclusive. .0.2 . . . . Right!

    (b) A and B are independent.

    (c) P(B|A) = 0.3

    (b) If A and B are independent: . P(A \cap B) \:=\:P(A)\!\cdot\!P(B) \:=\:(0.5)(0.7) \:=\:0.35

    Substitute into: . P(A \cup B) \:=\:P(A) + P(B) - P(A \cap B)

    . . . . . . . . . . . . . . . 0.7 \quad=\;0.5 \;+ \;P(B)\quad -\quad 0.35

    . . . . . . . . . . . . . . P(B) \;=\;0.7 - 0.5 + 0.35 \;=\;\boxed{0.55}


    (c) P(B|A) = 0.3

    From Bayes' Theorem: . P(B|A) \:=\:\frac{P(B \cap A)}{P(A)} \:=\:0.3

    So we have: . \frac{P(A \cap B)}{0.5} \:=\:0.3 \quad\Rightarrow\quad P(A \cap B) \:=\:(0.3)(0.5) \:=\:0.15

    Substitute into: . P(A \cup B) \:=\:P(A) + P(B) - P(A \cap B)

    . . . . . . . . . . . . . . . 0.7 \quad=\;0.5 \;+ \;P(B)\quad -\quad 0.15

    . . . . . . . . . . . . . . P(B) \;=\;0.7 - 0.5 + 0.15 \;=\;\boxed{0.35}

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