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Probability problem help

  1. Jun 2, 2012 #1
    Problem:

    A bag contains 50 coins. One of the coins has heads on both sides. 6 coins are drawn from the bag one by one at random without replacement. Each of the first 5 coins drawn is then tossed. What are the probabilities that the sixth coin drawn is the coin with both heads if 5 heads turn up in the 5 tosses?

    My Attempts:
    1st:
    The required probability
    =P(6th coin both heads | 5 heads turn up)
    =[p(6th coin both heads [itex]\cap[/itex]5 heads turn up)] /[P(5 heads turn up)]
    ={[(49P5)/(50P5)][1/(2^5)](1/45)}/{[1-(49P5)/(50P5)][1/(2^4)]+[(49P5)/(50P5)][1/(2^5)]}
    =1/55

    2nd:
    The required probability
    =[P(6th coin both heads |specified coin not drawn in first 5 draws)][P(specified coin not drawn in first 5 draws)]
    =(1/45)[(49P5)/(50P5)]
    =1/50

    Questions:
    Is any of my attempts correct?/Which 1 is correct?/What did I miss?/What should be the correct solution?/The problem not good? Please explain?
     
  2. jcsd
  3. Jun 2, 2012 #2
    The 2nd attempt doesn't make sense to me.. The event you want to condition on is that 5 heads turn up. So the first two lines of your first attempt make sense.

    Can you explain your steps on the third line?
     
  4. Jun 2, 2012 #3
    Assuming you haven't drawn the double headed coin in the first five draws, your probability of drawing it on the sixth draw is 1/45. Then you must determine the probability that you haven't drawn the coin in the first five draws and multiply this by 1/45. Do you know how to do this?
     
  5. Jun 2, 2012 #4
    Combinatorics; as fun as dangerous. Here goes my two cents:

    OK, in your first solution you have the right argument with the wrong answer and in your second solution you have the wrong argument with the right answer though calculated the wrong way too.

    The answer is 1/50, if you have five heads already you don't know if any of those coins is the one with two heads of if it is still ahead to come, so having 5 heads is like having no information at all, just like if they tell to you that you have 49 heads and then they ask you what is the chance the next coin is going to be the one with two heads; it would still be 1/50.
     
    Last edited: Jun 2, 2012
  6. Jun 2, 2012 #5
    So, if I draw 49 coins and ask what is the probability of drawing the double headed coin on the 50th draw given I haven't drawn the coin yet, you say it's 1/50?

    EDIT: As I read the question, the results of the coin tosses have nothing to do with the question, which as I read it, is what is the probability that the sixth draw draw will be the double headed coin. Since there is only one double headed coin, by definition the question assumes the first five coins can't be the double headed coin.
     
    Last edited: Jun 2, 2012
  7. Jun 2, 2012 #6
    You don't know if you haven't drawn the double headed coin yet, that's the point.

    The questions says nowhere that the first five coins can't be the double headed. In fact, if the problem implied that, then would make no sense to insist in saying that the first five coins are heads, if they are not the double headed it would not matter if they are heads or tails... So I think that the problem says what the problem says and that any of the first 5 coins could be the double headed.

    I believe actually that the whole point of the problem is to trick you... which makes it nice in that sense.
     
    Last edited: Jun 2, 2012
  8. Jun 2, 2012 #7
    The question is "A bag contains 50 coins. One of the coins has heads on both sides. 6 coins are drawn from the bag one by one at random without replacement. Each of the first 5 coins drawn is then tossed. What are the probabilities that the sixth coin drawn is the coin with both heads if 5 heads turn up in the 5 tosses?"

    It asks specifically for the probability that the sixth coin in sequence is the double headed coin. It doesn't matter that he tossed the previous coins and they came up heads (or whatever). None of the first five coins could be the double headed coin if the sixth coin is. (see post 3). Without replacement the probability that the sixth is doubleheaded is 1/45 if the none of the first five coins are. You calculate that probability as (49/50)(48/49).......(45/46) and multiply it by 1/45.

    I'm frankly tired of this nonsense. In an earlier thread the question clearly stated one die was observed to be a six. What does the word 'observed' mean to you?. To me it means that the probability that the other die is 6 is 1/6 in which case both dice show sixes.
     
    Last edited: Jun 2, 2012
  9. Jun 3, 2012 #8
    Calm down, you need to learn to disagree with people in a polite way. If I remember correctly in the thread you are talking about I was not the only one disagreeing with you anyway.

    The fun thing about forums is being able to disagree and discuss ideas, behaving like if you are in possession of the truth and those disagreeing with you just say "nonsense" is not a good starting point.

    IMHO some users in PF just take thing too seriously and they just miss the fun.
     
  10. Jun 3, 2012 #9
    I was working with the OP to arrive at the answer by a mathematical process and asked if he or she could solve the second part of the process when you posted the answer. Yes, the result is 1/50 when you solve the problem this way. However I find it very annoying since, in addition, you gave no process for understanding how this result is obtained. As for the other thread, DH was the last poster and he agreed that if you observe one die to be a 6, then the answer is that the remaining die is 6 with probability 1/6. That thread was derailed because some posters including you chose to solve a different problem.
     
    Last edited: Jun 3, 2012
  11. Jun 3, 2012 #10
    I kindly disagree... again :smile:
     
  12. Jun 3, 2012 #11
    Explain.
     
  13. Jun 3, 2012 #12
    I already did, for me the problem tricks you into thinking that it is giving you information when actually it is giving you none; any coin could be the double headed and, therefore, the probability is 1/50, it would be 1/50 if they ask you if it could be the first coin, the second coin... or any other coin. The fact that they show heads is like if they show nothing, like if they would stay in the bag and they ask you what is the chance you get it first, so there is no a mathematical process to reach 1/50 beyond the fact that there are 50 choices and only 1 is right.

    So that is my point of view, you have a different one and that is OK. Sometimes I'll be right, sometimes I'll be wrong and sometimes maybe both at the same time, but it is good people see different points of view, if anything even just to realize how many different approaches people have when dealing with a problem.

    In other forums people delete your answer if a number of users don't agree with you, I wouldn't delete yours even if I think is not the right one because, after all, discussions is what a forum is for; I just will present my point, you present yours, other people will do likewise and everyone can enjoy the discussion.

    So don't take it to a personal level, or if I was trying to annoy you for some reason. Maybe you could try to figure out why I believe I am right; the discussion could be fruitful. If you become annoyed by people holding different views then you can never enjoy a forum.

    That is all.
     
    Last edited: Jun 3, 2012
  14. Jun 3, 2012 #13
    You guys seem to be getting upset with each other. The answer is 1/50. You have no information to exclude the two-sided coin from the first five tossed because they are all heads. So, given the information that you have, the two sided coin is equally likely to be any of the fifty coins. I like to think of the extreme. If I lined up all fifty coins, heads up, then you have no information to exclude any coin. If, however, any of the first five were tails, then the problem changes as you have further information to exclude the first five. You could solve this using conditional probabilities ut the calculations are nasty.
     
  15. Jun 3, 2012 #14
    Actually not nasty, I misspoke.

    Let A = event that 6th coin is two headed

    Let B = event that two headed coin is one of the first five

    P(A) = P(A/B)P(B) + P(A/not B){1-P(B)}

    = (0)(1/10) + (1/45)(9/10)

    = 1/50
     
  16. Jun 3, 2012 #15

    berkeman

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    Staff: Mentor

    Question answered. Thread closed.
     
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