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Math Help - Probability help (this is driving me nuts) #3

  1. #1
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    Probability help (this is driving me nuts) #3

    I have been at this for 3 days already and i did some of them but i just am so confused that my brain hurts even trying to figure out if i am doing the right stuff. please please help me/ show me or point me in the right direction with these problems

    1.English and American spellings are colour and color, respectively. A man staying at a
    Parisian hotel writes this word, and a letter taken at random from his spelling is found to be a vowel. If
    40 percent of the English-speaking men at the hotel are English and 60 percent are Americans, what is the
    probability that the writer is an Englishman?


    4. Three missiles are fired at a target and hit it independently with probabilities 0.6, 0.7, and
    0.8, respectively. What is the probability that the target is hit?
    Last edited by mr fantastic; September 17th 2009 at 04:41 AM. Reason: Questions moved from another thread.
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  2. #2
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    Hello, SydneyBristow!

    Here's the first one . . .


    1. English and American spellings are "colour" and "color", respectively.
    A man staying at a Parisian hotel writes this word, and a letter taken
    at random from his spelling is found to be a vowel.
    If 40% of the English-speaking men at the hotel are English and 60% are Americans,
    what is the probability that the writer is an Englishman?
    This requires repeated use of Bayes' Theorem . . .


    We have: . \begin{array}{cccc}P(\text{UK}) \:=\: 0.4 & & P(\text{vowel}\,|\,\text{UK}) \:=\: \frac{3}{6} \:=\:0.5  & {\color{blue}[1]} \\ \\[-4mm] P(\text{US}) \:=\:0.6 & & P(\text{vowel}\,|\,\text{US}) \:=\:\frac{2}{5} \:=\:0.4& {\color{blue}[2]}\end{array}

    We want: . P(\text{UK}\,|\,\text{vowel}) \;=\;\frac{P(\text{UK} \wedge\text{vowel})}{P(\text{vowel})} \;\;{\color{blue}[3]}


    From [1] we have: . \frac{P(\text{vowel}\wedge\text{UK})}{P(\text{UK})  } \:=\:0.5 \quad\Rightarrow\quad \frac{P(\text{vowel}\wedge\text{UK})}{0.4} \:=\:0.5

    . .  P(\text{vowel}\wedge\text{UK}) \:=\:(0.4)(0.5) \:=\:0.20\;\;{\color{blue}[4]}

    From [2] we have: . \frac{P(\text{vowel}\wedge\text{US})}{P(\text{US})  } \;=\; 0.4 \quad\Rightarrow\quad \frac{P(\text{vowel}\wedge\text{US})}{0.6} \;=\;0.4

    . . P(\text{vowel}\wedge\text{US}) \;=\;(0.6)(0.4) \;=\;0.24


    Hence: . P(\text{vowel}) \;=\;P(\text{vowel}\wedge\text{UK}) + P(\text{vowel}\wedge\text{US}) \;=\;0.20 + 0.24 \;=\;0.44\;\;{\color{blue}[5]}



    Substitute [4] and [5] into [3]: .

    . . P(\text{UK}\,|\,\text{vowel}) \;=\;\frac{P(\text{UK}\wedge\text{vowel})}{P(\text  {vowel})} \;=\;\frac{0.20}{0.44} \;=\;\frac{5}{11}

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  3. #3
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    thanks so much

    thankyou .

    can you plz advise if for the missle problem my logic accurate?


    1-(.4*.3*.2 )=.976


    i cant type much as i hv newborn in my arns
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  4. #4
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    Probability that at least 1 hits = 1-Probability that every one misses

    =1-(.4*.3*.2) as you said, looks good to me
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