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Math Help - need a quick hand in basic probability

  1. #1
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    need a quick hand in basic probability

    Need a quick hand with basic probability

    Hi. I知 on a crash self-taught course in statistics, in preparation for a masters programme I値l be starting in a few months. I am using a book designed to help high school students prepare for state exams. The book痴 initial explanations don稚 always seem to prepare you for the exercises which it then throws at you. It gives solutions, but not worked solutions. I知 reasonably numerate but for some reason am stumped by a few of the basic questions on estimating probability. I need a boost.

    I learn well by studying worked examples (I recently re-learned all my high school algebra in a week this way). I知 wondering if someone would be kind enough to work out the following few problems for me, showing all the steps. You should find them easy. (I should be embarrassed, but am too intent on trying to learn this stuff to shy away from asking 電umb questions.)

    --------------

    Background to the questions:

    The following rules have already been spelled out in the text. They are:

    1) For any events a and b,
    p(NOT a) = 1 p(a)

    2) For any events a and b,
    p(a OR b) = p(a) + p(b) p(a AND b)

    3) For any mutually exclusive events a and b,
    p(a OR b) = p(a) + p(b)

    Later, in the course of explaining an answer to a problem involving 2 non-mutually exclusive events a and b, the book also states that

    p ( one or the other) = p( a OR b) p( a AND b)

    (They also use a Venn diagram or two, which I find easy to understand when they are completed but for some reason difficult to construct myself. At all events, I壇 prefer to learn to apply the above rules)

    ---------------------

    As for the problematic questions -- I seem to be getting stuck mainly on questions involving the NOT condition. The questions are:

    1) 典he events a and b are such that p(a) = 0.65, p(b) = 0.6 and p( a OR b) = 0.85.

    擢ind

    a) p(NOT a) [no problem here it痴 0.35]

    b) p (a AND b) [ again, no problem using rule 2) above, it痴 0.40]

    c) p (a AND (NOT b)) [here is where I get stuck they give the answer as 0.25, but I don稚 see how they reached it. Maybe fixating on the rules is somehow subverting my common sense.]

    --------------------------

    2) 典he events R and S are such that p(r) = 0.4 ; p (r OR s) = 0.7; and p(r AND S) = 0.2.

    擢ind

    a) p (NOT r) [ I see why it痴 0.6]

    b) p (S) [I see why, in light of the givens and rule 2), the answer is 0.5]

    c) p ( r AND (NOT s)) [ I don稚 understand their given answer of 0.2]

    ---------------

    3) The events a and b are such that p(a AND (NOT B)) = 0.37; p ((NOT a) AND b) = 0.41; and p((NOT a) AND (NOT b)) = 0.13.

    擢ind

    a) p (a AND b)

    b) p (a)

    c) p (NOT a) [If I could reach the answer to part b) I could reach the answer to this]

    d) p ( a OR b)

    They give the answers as a) 0.09 ; b) 0.46 ; c) 0. 54 [which I could have gotten if I were able to get b) ] and d) 0.87 [ which I could reach if I could reach the answers to parts a) and b) and somehow work out the positive value of the value b.]

    What is it that I知 missing?

    --------------
    The last one

    4) 鉄tudents must study at least 1 of the following: French (F), German (G) or Spanish (S).

    65% study F; 57% study G; and 36% study S.
    40% study F and G. 13% study F and S.
    5% study all 3.

    擢ind the probability that a student picked at random studies

    i) only F ii) S but not G.

    [Must admit I知 not sure where to begin]

    I壇 be grateful for any help, as I will not be able to go on to the sections on conditional probability and independent events until I am firm on these basics.

    Sincerely,

    Ken
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by lingyai

    1) 典he events a and b are such that p(a) = 0.65, p(b) = 0.6 and p( a OR b) = 0.85.

    擢ind

    a) p(NOT a) [no problem here it痴 0.35]

    b) p (a AND b) [ again, no problem using rule 2) above, it痴 0.40]

    c) p (a AND (NOT b)) [here is where I get stuck they give the answer as 0.25, but I don稚 see how they reached it. Maybe fixating on the rules is somehow subverting my common sense.]

    For c) you need to observe that:

    p(a AND b)+p(a AND (NOT b))=p(a)=1-p(NOT a),

    two of these terms have already been established in parts a) and b).
    Plug these in and you will find:

    0.4+p(a AND (NOT b))=0.65,

    which will give the given solution.

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by lingyai

    2) 典he events R and S are such that p(r) = 0.4 ; p (r OR s) = 0.7; and p(r AND S) = 0.2.

    擢ind

    a) p (NOT r) [ I see why it痴 0.6]

    b) p (S) [I see why, in light of the givens and rule 2), the answer is 0.5]

    c) p ( r AND (NOT s)) [ I don稚 understand their given answer of 0.2]

    ---------------
    Observe:

    p(r AND s)+p(r AND (NOT s))=p(r),

    plug in the givens:

    0.2+p(r AND (NOT s))=0.4,

    andthe result for c) follows.

    RonL
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by lingyai
    3) The events a and b are such that p(a AND (NOT B)) = 0.37; p ((NOT a) AND b) = 0.41; and p((NOT a) AND (NOT b)) = 0.13.

    擢ind

    a) p (a AND b)

    b) p (a)

    c) p (NOT a) [If I could reach the answer to part b) I could reach the answer to this]

    d) p ( a OR b)

    They give the answers as a) 0.09 ; b) 0.46 ; c) 0. 54 [which I could have gotten if I were able to get b) ] and d) 0.87 [ which I could reach if I could reach the answers to parts a) and b) and somehow work out the positive value of the value b.]

    What is it that I知 missing?

    --------------
    Observe:

    p ((NOT a) AND b) + p((NOT a) AND (NOT b)) = p(NOT a)=1-p(a).

    The terms on the left hand side are given so p(a) may be found:

    0.41+0.13=0.54=1-p(a),

    so p(a)=0.46

    RonL
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by lingyai
    --------------
    The last one

    4) 鉄tudents must study at least 1 of the following: French (F), German (G) or Spanish (S).

    65% study F; 57% study G; and 36% study S.
    40% study F and G. 13% study F and S.
    5% study all 3.

    擢ind the probability that a student picked at random studies

    i) only F ii) S but not G.
    i) 40% study F&G, but 5% study all 3, so 35% study only F&G.
    13% study F&S, but 5% study all 3, so 8% study only F&S.

    So the proportion that studies only F is:65-35-8-5=17%.

    ii) I'm not sure there is sufficient information to do this.

    RonL
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