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Math Help - Some probability help again please...

  1. #1
    Bar0n janvdl's Avatar
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    Some probability help again please...

    Hey guys, just 3 questions this time...

    1. Three different machines M1, M2, and M3 were used for producing a large
    batch of similar manufactured items. Suppose that 20% of the items were
    produced by M1, 30% by M2, and 50% by M3. Suppose further that 1% of the items produced by M1 are defective, that 2% of the items produced by M2 are defective, and that 3% of the items produced by M3 are defective.

    i) Suppose one item is selected at random from the entire batch and it
    is found to be defective. Determine the probability that this item
    was produced by M2.

    ii) Suppose one item is selected at random from the entire batch and it
    is found to be nondefective. Determine the probability that it was
    produced by M2.

    My answers:

    We must use conditional probability.

    i)

    Code:
                               M1             M2            M3
    Items Produced (Ip)  :    0,2            0,3           0,5
    Items Defective (Id) :   0,01           0,02         0,03
    Items not-Defect. (In):  0,99           0,98         0,97
    P( M2 \cap Id) = P(M2) \cdot P(Id | M2)

    = 0,3 \times 0,02

    = 0,006

    = 0,6%


    ii) P(M2 \cap In) = P(M2) \cdot P(In | M2)

    = 0,3 \times 0,98

    = 0,294

    = 29,4%



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


    2. A box contains 3 coins with head on each side, 4 coins with a tail on each side, and 2 fair coins. One of these 9 coins is selected at random and tossed once. Suppose a head is obtained. Determine the probability that the coin is one with head on each side.

    My answer:

    We obtain a head so we know that it is not one of the coins with the tail on each side. So our coin will be one out of the 3 head-only coins and the 2 fair coins.

    We have 5 coins and 3 of them have two heads.

    So the probability is 3 out of 5 equals 60%.



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



    3. In a certain city, 30% of the people are Conservatives, 50% are Liberals, and 20% are Independents. Records show that in a particular election, 65% of the Conservatives voted, 82% of the liberals voted, and 50% of the Independents voted. If a person in the city is selected at random and it is learned that he did not vote in the last election, what is the probability that he is a Liberal?

    My answer:

    We must use conditional probability.

    Code:
                           Consv         Lib         Ind
    Population:            0,3             0,5        0,2
    Voted(V):             0,65            0,82        0,5
    Not Voted(N):         0,35            0,18        0,5
    P(Lib \cap N) = P(Lib) \cdot P(N | Lib)

    = 0,5 \times 0,18

    = 0,09

    = 9%
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  2. #2
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    galactus's Avatar
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    Something I do with these problems is to make a chart by assuming a certain amount of items produced. 1000 is a nice number to work with.

    That means we have:


    Code:
                       M1                M2                M3         total
    ------------------------------------------------------
    
    good            198                294               485         977
    
    
    
    bad               2                   6                   15           23       
    
    -----------------------------------------------------------------
                      200               300                 500            1000
    So, probability it's a defect given it came from M2 would be 6/300=.02

    Probability it came from M2 given it's a defect is 6/23=0.26

    and so forth.

    This makes things easier most times.
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  3. #3
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    Quote Originally Posted by janvdl View Post
    Hey guys, just 3 questions this time...

    1. Three different machines M1, M2, and M3 were used for producing a large
    batch of similar manufactured items. Suppose that 20% of the items were
    produced by M1, 30% by M2, and 50% by M3. Suppose further that 1% of the items produced by M1 are defective, that 2% of the items produced by M2 are defective, and that 3% of the items produced by M3 are defective.

    i) Suppose one item is selected at random from the entire batch and it
    is found to be defective. Determine the probability that this item
    was produced by M2.

    ii) Suppose one item is selected at random from the entire batch and it
    is found to be nondefective. Determine the probability that it was
    produced by M2.

    My answers:

    We must use conditional probability.

    i)

    Code:
                               M1             M2            M3
    Items Produced (Ip)  :    0,2            0,3           0,5
    Items Defective (Id) :   0,01           0,02         0,03
    Items not-Defect. (In):  0,99           0,98         0,97
    P( M2 \cap Id) = P(M2) \cdot P(Id | M2)

    = 0,3 \times 0,02

    = 0,006

    = 0,6%


    ii) P(M2 \cap In) = P(M2) \cdot P(In | M2)

    = 0,3 \times 0,98

    = 0,294

    = 29,4%
    [snip]
    i) You want P(M2 | Id) = \frac{P( M2 \cap Id)}{P(Id)}.

    You already have P( M2 \cap Id) = 0,006.

    P(Id) = P(Id|M1) \cdot P(M1) + P(Id|M2) \cdot P(M2) + P(Id|M3) \cdot P(M3) = ......


    ii) You want P(M2 | In) = \frac{P( M2 \cap In)}{P(In)}.

    You already have P( M2 \cap In) = 0,294.

    P(In) = ......
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  4. #4
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    Quote Originally Posted by janvdl View Post
    Hey guys, just 3 questions this time...

    [snip]
    2. A box contains 3 coins with head on each side, 4 coins with a tail on each side, and 2 fair coins. One of these 9 coins is selected at random and tossed once. Suppose a head is obtained. Determine the probability that the coin is one with head on each side.

    My answer:

    We obtain a head so we know that it is not one of the coins with the tail on each side. So our coin will be one out of the 3 head-only coins and the 2 fair coins.

    We have 5 coins and 3 of them have two heads.

    So the probability is 3 out of 5 equals 60%.
    [snip]
    A tree diagram is useful here. Draw one. Use it to get the following answer:

    \frac{\frac{3}{9}}{\frac{3}{9} + 0 + \frac{1}{9}} = \frac{3}{4} ....
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  5. #5
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    Quote Originally Posted by janvdl View Post
    Hey guys, just 3 questions this time...

    [snip]
    3. In a certain city, 30% of the people are Conservatives, 50% are Liberals, and 20% are Independents. Records show that in a particular election, 65% of the Conservatives voted, 82% of the liberals voted, and 50% of the Independents voted. If a person in the city is selected at random and it is learned that he did not vote in the last election, what is the probability that he is a Liberal?

    My answer:

    We must use conditional probability.

    Code:
                           Consv         Lib         Ind
    Population:            0,3             0,5        0,2
    Voted(V):             0,65            0,82        0,5
    Not Voted(N):         0,35            0,18        0,5
    P(Lib \cap N) = P(Lib) \cdot P(N | Lib)

    = 0,5 \times 0,18

    = 0,09

    = 9%
    You want P(Lib | N).

    It'll follow the same process as Q1 ......

    By the way, Galactus has given an excellent approch to use. I was too lazy to suggest it
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  6. #6
    Bar0n janvdl's Avatar
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    Thanks for the help guys... I'm lost on this though.
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  7. #7
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    Quote Originally Posted by janvdl View Post
    Thanks for the help guys... I'm lost on this though.
    These particular questions or the topic in general? If the former, where do you lose it? If the latter, stick at it - it will click eventually. Just keep doing questions until it does.

    Galactus gave excellent advice. Tree diagrams can also help in visualising what's going on. I usually shy away from the formula approach as it's easy to get lost.
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  8. #8
    Bar0n janvdl's Avatar
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    Quote Originally Posted by mr fantastic View Post
    These particular questions or the topic in general? If the former, where do you lose it? If the latter, stick at it - it will click eventually. Just keep doing questions until it does.

    Galactus gave excellent advice. Tree diagrams can also help in visualising what's going on. I usually shy away from the formula approach as it's easy to get lost.
    The topic in general. Yeah i guess I'll get the hang of it soon. I just don't know when to use factorials or Bayes' Rule or Conditional Probability... It's just very confusing.

    I'll have to google tree diagrams. I have no idea what that is.
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