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Some probability questions - mean , variance , uniformly chosen points

  1. Nov 2, 2013 #1
    Hi there ,
    I have some simple (on 1st look) questions. I just need some help.http://img59.imageshack.us/img59/3018/5hby.png [Broken]
    1st question.
    I think the expected value should be 150 since the probability of getting a head is 1/2 .. 300x1/2 = 150 so the mean should also be 150.
    the variance i think is 1/2x1/2x300 = 75
    2nd question.
    Here I have something for 6 eggs with 2 bad and 4 good.. but when I use the same logic for 600 eggs total, 20 bad I get very big number. So my logic for 6 eggs 4 good 2 bad is:
    Let X be the number of bad eggs picked. If X=0, we have 4/6x3/5= 6/15 =2/5
    P(X-1)= 4/6x2/5 + 2/6x4/5 =8/15
    P(X=2) = 1/15 so the EX is 0x6/15+1x8/15+2x1/15 = 2/3
    But how I can use this logic for the bigger problem?
    3rd question
    Why do we have to care about the 1st line? Since the probability of picking one of the 2 coins is 1/2.
    a) 2 heads P(HHT)+P(THH)+P(HTH)=3/8?
    b)if the 1st one is heads so we need one head out of 2 flips so isnt it just 1/2?
    or I am missing something here?
    4th question
    I think it will be better to start from part b)
    b) I think the length function is f(x)=min(x,1-x) as x is on the interval [0,0.5] and 1-x is on the interval [0.5,1], We have to find E(f(X)) which is equal to:
    http://img707.imageshack.us/img707/1160/3taz.jpg [Broken]
    a) from part b) since the average size is 1/4 it can be either left or right segment. So the left can be either 1/4 or 3/4
    c) don't have idea about this question.. intuitively think something about 2/3
    but cannot explain why :D
    Thanks
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 2, 2013 #2

    cepheid

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    (1) Yes your solution is correct, but do you know why?

    (2) You have a 1/30 chance of picking a bad egg. So if you pick 20 eggs, what is the expected number of bad ones?

    (3) You need to use conditional probability here, since the probability of getting two heads is *different* depending on which of the coins you picked.
     
  4. Nov 2, 2013 #3
    2)20/30?
    Why the probability of getting haed is dependent of the coin you pick if they are equally probable? maybe just have to add 1/2 to the chance of getting head?
    any thoughts for 4th question?
     
    Last edited: Nov 2, 2013
  5. Nov 2, 2013 #4

    cepheid

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    Yes, 20/30.

    Suppose I pick the coin that has a 2/3 chance of showing up heads. The probability that the outcome of three tosses is that two of them are heads is different for this coin vs. for the fair. Coin. You can think of the former probability as the conditional probability of getting two heads GIVEN that you pick the unfair coin. The latter probability is the conditional probability of getting two heads GIVEN that you pick the fair coin. You need to combine these two using the knowledge of how likely it is that you pick one or the other. I would recommend brushing up on conditional probability for this type of problem.
     
  6. Nov 2, 2013 #5

    Ray Vickson

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    For question 3: P{H=2} = P{H=2|coin A}*P{coin A} + P{H=2|coin B}*P{coin B}.
     
  7. Nov 2, 2013 #6

    D H

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    I wouldn't have phrased it this way. Stating it as a 1/30 chance of picking a bad egg implies sampling with replacement. This is a sampling without replacement problem.

    There's a rather interesting relationship between the mean for a sampling with replacement problem and a sampling without replacement problem.

    I agree. This third problem is all about conditional probabilities.


    Regarding the fourth problem, you're overthinking it, gl0ck. Start with part (a) first.
     
  8. Nov 2, 2013 #7
    @DH
    Then the 2nd problem?
    2) P(n bad eggs) = C(20,n)*C(580,20-n)/C(600,20)
    The sum of the products is 2/3. ?

    3a) P(2 heads with coin 1) = 3 * 1/8 = 3/8
    P(2 heads with coin 2) = 3 * 4/27 = 4/9
    P(2 heads) = (1/2 * 3/8) + (1/2 * 4/9) = 59/144

    3b) P(you have the first coin) = (1/2) / (1/2 + 2/3) = 3/7
    P(you have the second coin) = (2/3) / (1/2 + 2/3) = 4/7
    If you have the first coin, P(1 more head) = 1/2
    If you have the second coin, P(1 more head) = 4/9
    P(2 heads) = (3/7 * 1/2) + (4/7 * 4/9) = 59/126

    4a) 1/2 (must be the same as the expected length of the right segment)
    4c) 2/3 - the break can occur in these regions: (0-1/3] and [2/3, 1)

    is this now correct?
     
  9. Nov 2, 2013 #8

    haruspex

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    A graph might help. Let x be the distance of the break from one end. Plot the length of the longer segment against x, likewise for the shorter segment. For what ranges of x is the ratio at least 2:1?
     
  10. Nov 2, 2013 #9

    D H

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    How did you obtain that "P(n bad eggs) = C(20,n)*C(580,20-n)/C(600,20)"? This is correct. You must have assumed some distribution to obtain that correct result.

    What is the mean for that distribution? Don't guess. You should be able to look it up.


    You have all parts of (3) and (4) correct. Congrats!
     
  11. Nov 2, 2013 #10
    P(n bad eggs) = C(20,n)*C(580,20-n)/C(600,20)

    There are C(20,n) ways to choose n bad eggs. There are C(580,20-n) ways to choose such that the rest of the eggs are good. Multiply these to get the number of ways to choose exactly n bad eggs. Divide by C(600,20) to get the probability of choosing exactly n bad eggs.

    To compute the expected number of bad eggs:
    sum(n = 0 to 20, n*P(n))
     
  12. Nov 2, 2013 #11

    D H

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    Right. So what probability distribution is that? Did you develop that formula for P(n) on your own, or did you look it up somewhere? If it's the latter, just look up the mean.

    If it's the former, that's very good on your part, and that means you should be able to calculate that sum.
     
  13. Nov 2, 2013 #12
    Just saw similar problem in one book.. Thanks very much guys
     
  14. Nov 2, 2013 #13

    haruspex

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    True, but it's a lot easier by induction on the number of selections.
     
  15. Nov 2, 2013 #14

    Ray Vickson

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    As others have said, congratulations on obtaining this (although you still need to actually do the sum). However, if all you want are expected values (rather than distributions or higher moments such as variance, etc.) you can use a much simpler approach---an approach that occurs over and over again and would make a useful part of your toolkit.

    The number of bad eggs is ##N = \sum_{i=1}^{20} X_i##, where ##X_1 = 1## if egg i is bad and ##X_i = 0## if egg i is good. Basically, the X_i just count the bad eggs. Now, of course, ##(X_1,X_2,\ldots,X_{20})## has a complicated multivariate distribution, but all we want is the expected sum ##EN = E \sum_{i} X_i = \sum_{i} E X_i##, and that is quite easy to get. The point is that in the last expression we need only the marginal distribution of ##X_i##, so all we need know is ##E X_i = P\{ \text{egg }i\text{ is bad} \}##. Here, we make no reference at all to any egg other than i, and for any i all the events {egg i is bad} have the same probability!

    Perhaps surprisingly, this would be true even if we drew out all 200 eggs: if I don't tell you the outcome on any of the first 199, the probability that the 200th egg will be bad is no different than the probability the first egg is bad! If you don't believe this, think about a sample-space representation: imagine labelling the eggs from 1 to 200, of which 20 are bad. Take a permutation of the 200 numbers and just look at the the event B_i ={egg in position i is bad}. There are the same number of permutations for B_1 or for B_2 or for B_3 ... or for B_200. Of course, P{egg i is bad} is the number of permutations in B_i, divided by the total number (200!) of permutations.
     
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