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Conditional probablility. HELP! HURRY!

  1. May 2, 2005 #1
    I missed my data management class on friday and now If find out I have a quiz tomorrow on the stuff I missed. I got fridays sheet but these questions seem really dumb and kind of confusing. Some of them seem so simple they are confusing.
    here are some of the ones that confused me that maybe you guys can help out with
    1) two dice are rolled . What is the probablility that the sum is 8 and the sum is even?
    This seems kind of redundant to me. I mean if the sum is 8 then the sum is obviously even, so why bother saying that the sum is even? very confusing
    2) You play a game of rilling two dice. You win if you roll a five before a seven. you continue to roll until you win or lose. what is probability that you win.
    ok so the probablility that you roll a five is 4/36 and of rolling a seven is 6/36 but rolling a 5 before a 7. what do I do with that.
    3) Chocolate bars are advertised to weigh 120g. It is found that there is a 65% probablility that the bar weighs 120g and a 20% probablility that it weighs less than 120g. what is the probablility that a bar, not weighing 120g is greater than 120g
    It seems obvious that the answer is 15 chance. I have a feeling that this is trying to trick me in some way. (I just wish I knew which way it was)
    4) a car manufacturing plant has 3 shifts working on the assembly line. THe morning shift produces 38% of the total production, the afternoon 34% and the evening 28%. Of their output 3%,2% and 1%, respectively, do not pass quality control. If a vehicle is selected randomly and found defective what is the probablilty that is was made by the a)morning b) afternoon c)evening shift.
    This one I really have no idea on. obviously I only need help with one section a,b or c. I could figure the rest out for myself from that.

    Well thanks for the help guys. I really need to ace this test tomorrow or my mark is gonna bomb.
     
  2. jcsd
  3. May 2, 2005 #2

    StatusX

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    1) Yes, it's redundant. Do you know how to do it?

    2) Ignore all the rolls where the dice doesn't come up 5 or 7. Out of this set, what is the probability of picking a 5 and what is it of picking a 7?

    3) I don't think it is a trick, it's just odd that there would be a finite chance of picking a specific value of a continuous variable. They must mean some range around 120, but I don't think you need to worry about that.

    4) Nothing tricky here. How many total defective cars are there, and how many does each shift make?
     
  4. May 2, 2005 #3

    OlderDan

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    Just look at #1 for now. I think the point of the first question is to make sure you understand what this means:

    P(A and B) = P(A)P(B|A)

    P(8 and E) = P(8)P(E|8) = P(8) by what is obvious to you.

    P(E and 8) = P(E)P(8|E) = ?? Now this is a bit more work. Are they really the same?
     
  5. May 2, 2005 #4

    OlderDan

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    You look at all the ways you can roll a 5 before rolling a 7 with the associated probabilities of each and add them all together. It's not so bad. There are only infinitely many possibilites :smile: Let me use some shorthand, and call S the probability of rolling a 5 on any one roll, F the probability of rolling a 7 on any one roll, and N the probability of rolling any other number on any one roll. As soon as you "succeed" or "fail" the game is over. You need to find the probabilities by computing the following sequences of probabilities:

    S
    F
    NS
    NF
    NNS
    NNF
    NNNS
    NNNF
    NNNNS
    NNNNF

    etc

    Add together the probabilities of all sequences that end in S and all the sequences that end in F and you will have what you need. It is helpful to be able to sum a geometric series, but you don't really need to if you keep in mind that the sum of both possible terminations must be 1.
     
  6. May 2, 2005 #5

    StatusX

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    No offense dan, but that seems a lot harder than my suggestion. Do you think my method is wrong?
     
  7. May 2, 2005 #6

    OlderDan

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    My apologies. I honestly did not even see your post when I posted mine. I was working on the first one when you posted yours and when I sent mine in, since it arrived after yours mine was the most recent. When I was working on the second one I kept seeing my name last on the list, and thought no one else had posted. I will try to be more careful in the future.

    I can't say that yours is wrong, since we both arrive at the same conclusion. All those Ns in my approach divide out and reduce to your result. So I agree yours is simpler. There is something a bit unsettling to me about "ignoring" the indecisive trials, but it seems you are quite right.
     
  8. May 2, 2005 #7
    I don't think you have to ignore the indecisive trials necessarily. It's just the total number of indecisive trials before getting a 5 or 7 in the set of all possible outcomes is equal for a 5 or 7, so the weighted probability of those outcomes factor out of your calculation.
     
  9. May 2, 2005 #8

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    I'm just saying if you were to write down the rolls as they came up, and then cross out all the rolls that weren't 5 or 7, the distribution would be 4/10 5 and 6/10 7, so these are the probabilities each would come up first.
     
  10. May 3, 2005 #9

    OlderDan

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    You are correct. If you don't ignore the indecisives and you look at the probability of terminting the game with a 5 by every possible path you have

    [tex] P(5) = \frac{4}{{36}}\sum\limits_{n = 0}^\infty {\left[ {\frac{{26}}{{36}}} \right]} ^n = \frac{4}{{36}}\left[ {\frac{1}{{1 - {{26} \mathord{\left/
    {\vphantom {{26} {36}}} \right.
    \kern-\nulldelimiterspace} {36}}}}} \right] = \frac{4}{{10}} = \frac{2}{5} [/tex]

    and the probability of terminating with a 7 is

    [tex] P(7) = \frac{6}{{36}}\sum\limits_{n = 0}^\infty {\left[ {\frac{{26}}{{36}}} \right]} ^n = \frac{6}{{36}}\left[ {\frac{1}{{1 - {{26} \mathord{\left/
    {\vphantom {{26} {36}}} \right.
    \kern-\nulldelimiterspace} {36}}}}} \right] = \frac{6}{{10}} = \frac{3}{5} [/tex]

    But since you know that P(5) + P(7) = 1, there is no need to do those sums. They do factor out

    [tex] P(5) = \frac{{P(5)}}{{P(5) + P(7)}} = \frac{{\frac{4}{{36}}}}{{\frac{4}{{36}} + \frac{6}{{36}}}} = \frac{4}{{10}} = \frac{2}{5}
    \] [/tex]
     
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