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Math Help - Some probability problems

  1. #1
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    Some probability problems

    Hello everybody!

    I don't have right answers for the problems below, so I would really appreciate if someone could check my answers. I'm not sure if any of this is right.


    1. We have 25 products, 5 of which are bad. What is the probability of
    a) choosing one bad product?
    b) choosing (any) five products?
    c) choosing five products with exactly two bad products?
    (Products are chosen randomly.)

    a) \frac{5}{25}=\frac{1}{5}

    b) \frac{5}{25}=\frac{1}{5}

    c) \frac{20}{25} \cdot \frac{19}{24} \cdot \frac{18}{23} \cdot \frac{5}{22} \cdot \frac{4}{21}=0.0215


    2. We have 8 products, 3 of which are bad. What is the probability of choosing either BGG or GBG (B - bad product, G - good product)?

    \frac{3}{8} \cdot \frac{5}{7} \cdot \frac{4}{6} \cdot 2=0.3571


    3. We have 7 good and 4 bad products. What is the probability of
    a) choosing three good products?
    b) choosing exactly two good products and one bad product?
    c) choosing three products with at least one bad product?

    a) \frac{7}{11} \cdot \frac{6}{10} \cdot \frac{5}{9}=0.2121

    b) \frac{7}{11} \cdot \frac{6}{10} \cdot \frac{4}{9}=0.1697

    c) \frac{4}{11} \cdot \frac{7}{10} \cdot \frac{6}{9} +<br />
\frac{4}{11} \cdot \frac{3}{10} \cdot \frac{7}{9} +<br />
\frac{4}{11} \cdot \frac{3}{10} \cdot \frac{2}{9}=0.2788
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  2. #2
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    Quote Originally Posted by Dr. Jekyll View Post
    Hello everybody!

    I don't have right answers for the problems below, so I would really appreciate if someone could check my answers. I'm not sure if any of this is right.


    1. We have 25 products, 5 of which are bad. What is the probability of
    a) choosing one bad product?
    b) choosing (any) five products?
    c) choosing five products with exactly two bad products?
    (Products are chosen randomly.)

    a) \frac{5}{25}=\frac{1}{5}

    b) \frac{5}{25}=\frac{1}{5}

    c) \frac{20}{25} \cdot \frac{19}{24} \cdot \frac{18}{23} \cdot \frac{5}{22} \cdot \frac{4}{21}=0.0215
    a)&b) i believe it's correct.

    but c) \frac{5C2*20C3}{25C5} = 0.215 (3s.f.)
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  3. #3
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    Quote Originally Posted by Dr. Jekyll View Post
    Hello everybody!

    2. We have 8 products, 3 of which are bad. What is the probability of choosing either BGG or GBG (B - bad product, G - good product)?

    \frac{3}{8} \cdot \frac{5}{7} \cdot \frac{4}{6} \cdot 2=0.3571
    I believe your answer is right.

    but it's not times 2

    it should be \frac{3}{8} \cdot \frac{5}{7} \cdot \frac{4}{6} + \frac{5}{8} \cdot \frac{3}{7} \cdot \frac{4}{6}
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  4. #4
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    Quote Originally Posted by Dr. Jekyll View Post
    Hello everybody!

    3. We have 7 good and 4 bad products. What is the probability of
    a) choosing three good products?
    b) choosing exactly two good products and one bad product?
    c) choosing three products with at least one bad product?

    a) \frac{7}{11} \cdot \frac{6}{10} \cdot \frac{5}{9}=0.2121

    b) \frac{7}{11} \cdot \frac{6}{10} \cdot \frac{4}{9}=0.1697

    c) \frac{4}{11} \cdot \frac{7}{10} \cdot \frac{6}{9} +<br />
\frac{4}{11} \cdot \frac{3}{10} \cdot \frac{7}{9} +<br />
\frac{4}{11} \cdot \frac{3}{10} \cdot \frac{2}{9}=0.2788
    a) is correct.

    b) should be \frac{7C2*4C1}{11C3}

    c) should be 1-(Probability of no bad product)
    = 1-\frac{7C3}{11C3}
    =0.788(3s.f.)
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  5. #5
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    Quote Originally Posted by BabyMilo View Post
    I believe your answer is right.

    but it's not times 2

    it should be \frac{3}{8} \cdot \frac{5}{7} \cdot \frac{4}{6} + \frac{5}{8} \cdot \frac{3}{7} \cdot \frac{4}{6}
    Well, isn't it the same thing?

    Quote Originally Posted by BabyMilo View Post
    a) is correct.

    b) should be \frac{7C2*4C1}{11C3}

    c) should be 1-(Probability of no bad product)
    = 1-\frac{7C3}{11C3}
    =0.788(3s.f.)
    Of course! Order in which we choose the products is not relevant, that's why we use binomial coefficients (number of combinations), am I right?

    There may be some more questions. Thanks a lot!
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  6. #6
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    Quote Originally Posted by Dr. Jekyll View Post
    Hello everybody!

    I don't have right answers for the problems below, so I would really appreciate if someone could check my answers. I'm not sure if any of this is right.


    1. We have 25 products, 5 of which are bad. What is the probability of
    a) choosing one bad product?
    b) choosing (any) five products?
    c) choosing five products with exactly two bad products?
    (Products are chosen randomly.)

    a) \frac{5}{25}=\frac{1}{5}

    b) \frac{5}{25}=\frac{1}{5} This precedes part c) and is referring to the probability of picking a particular group of 5

    \color{blue}\frac{1}{\binom{25}{5}}

    c) \frac{20}{25} \cdot \frac{19}{24} \cdot \frac{18}{23} \cdot \frac{5}{22} \cdot \frac{4}{21}=0.0215 this is the probability that the last 2 picked were bad,

    which equals the probability of any 2 of the 5 being bad, those possibilities must be summed.



    2. We have 8 products, 3 of which are bad. What is the probability of choosing either BGG or GBG (B - bad product, G - good product)?

    \frac{3}{8} \cdot \frac{5}{7} \cdot \frac{4}{6} \cdot 2=0.3571


    3. We have 7 good and 4 bad products. What is the probability of
    a) choosing three good products?
    b) choosing exactly two good products and one bad product?
    c) choosing three products with at least one bad product?

    a) \frac{7}{11} \cdot \frac{6}{10} \cdot \frac{5}{9}=0.2121

    b) \frac{7}{11} \cdot \frac{6}{10} \cdot \frac{4}{9}=0.1697 this is the probability that only the last one chosen is bad,

    we need to sum the probabilities of only the 1st, only the 2nd or only the 3rd is bad.


    c) \frac{4}{11} \cdot \frac{7}{10} \cdot \frac{6}{9} +<br />
\frac{4}{11} \cdot \frac{3}{10} \cdot \frac{7}{9} +<br />
\frac{4}{11} \cdot \frac{3}{10} \cdot \frac{2}{9}=0.2788

    This is the probability that the 1st is bad or the 1st and 2nd are bad or all 3 are bad.
    It doesn't account for only the 2nd being bad, only the last being bad, only the last 2 being bad, the 1st and last being bad.
    Hi Dr Jekyll,

    some additional feedback on your work and why it's simplest to calculate
    as BabyMilo showed.
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  7. #7
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    3. We have 7 good and 4 bad products. What is the probability of
    a) choosing three good products?
    b) choosing exactly two good products and one bad product?
    c) choosing three products with at least one bad product?

    I've been looking at this problem 3. Maybe I'm wrong but wouldn't it be easiest to just use a hypergeometric distribution?

    For example:
    A)
    x= the number of good products
    X~hypergeometric(N=11, n=3, k=7)
    pr(X=3)

     \frac{(^{k}_{x})(^{N-k}_{n-x})}{(^{N}_{n})}

     \frac{(^{7}_{3})(^{4}_{0})}{(^{11}_{3})}\approx 0.2121

    B)
    x= the number of good products
    X~hypergeometric(N=11, n=3, k=7)
    pr(X=2)

     \frac{(^{k}_{x})(^{N-k}_{n-x})}{(^{N}_{n})}

     \frac{(^{7}_{2})(^{4}_{1})}{(^{11}_{3})}\approx 0.5091

    C) x= the number of bad products
    X~hypergeometric(N=11, n=3, k=4)

    pr(X \geq 1)= \sum ^{3}_{x=1} \frac{(^{k}_{x})(^{N-k}_{n-x})}{(^{N}_{n})} = \frac{(^{4}_{1})(^{7}_{2})}{(^{11}_{3})} + \frac{(^{4}_{2})(^{7}_{1})}{(^{11}_{3})} + \frac{(^{4}_{3})(^{7}_{0})}{(^{11}_{3})} \approx 0.7879

    Maybe I'm wrong, but I believe that is the proper way these problems. Any insight into what I'm doing right or wrong would be awesome.
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