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Math Help - Independent probability

  1. #1
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    Question Independent probability

    E and F are independent with P(E)=0.35 and P(F)=0.45. P(E union F)=___.
    ok this one I have some trouble with. P(E) - P(F) = 0.10 this is my answer is it right. this looks easy but hard for me. list of possible answers are.

    1. 0.80
    2. 0.10
    3. 0.1575
    d. 0.6425
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  2. #2
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    Cap'n, you're much more learned than me so correct me if I am wrong, but doesn't P(E union F) mean P(E or F), not P(E and F)?.

    Since E and F are independent, P(E)P(F)=(.35)(.45)=.1575

    P(E or F)=P(E)+P(F)-P(E or F)=.35+.45-.1575=.6425, assuming they're not mutually exclusive.

    Then again, mutually exclusive events can not be independent events.
    Last edited by galactus; August 8th 2006 at 02:39 PM.
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  3. #3
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    Question ok I see

    Ok I see both ways, but don't a P(E union F) mean it needs to be added. I only ask this because thats what my text book shows how to do it. So please let me know hich is right, and thanks so much again for all the help.
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  4. #4
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    Hello, kwtolley!

    I agree with Galactus . . .


    E and F are independent with: P(E)=0.35 and P(F)=0.45.\qquad P(E \cup F) = \_\_\_

    1)\;0.80\qquad 2)\;0.10\qquad 3)\;0.1575\qquad 4)\;0.6425

    Formula: . \boxed{P(E \cup F)\;=\;P(E) + P(F) - P(E \cap F)}


    Since E and F are independent: . 0.35)(0.45)\:=\:0.1575" alt="P(E \cap F) \:=\:P(E)\cdot P(F) \:=\0.35)(0.45)\:=\:0.1575" />

    Therefore: . P(E \cup F) \;=\;0.35 + 0.45 - 0.1575 \;= \;0.6425 . . . answer (d)

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  5. #5
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    Smile thanks for the help

    ok I got it
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