John Salerno Posted November 18, 2017 Share Posted November 18, 2017 Hello everyone. I downloaded a puzzle app from the Play Store that has probability puzzles, and I hate to admit I'm already stuck on the second puzzle! But I feel like I'm doing it correctly, though it says my answer is incorrect. I'm just looking for some general guidance for if I'm on the right track. I don't want the answer or a direct solution. If I'm wrong, a suggestion for certain keywords I could search for and read up on would be helpful too. Here is the puzzle: Quote Six individual socks are sitting in your drawer: two red, two blue, and two purple. It's dark -- you can't see a thing. You pick a first sock uniformly at random, and then a second sock from the remaining five. Assume the two choices are independent. What's the probability you end up with two socks of the same color? First off, I'm wondering if the word "uniformly" means something that I'm not aware of. I've never studied probability, so perhaps that's one aspect I'm missing. Second, even though it seems like drawing the second sock would be a dependent event, I don't understand why it says "Assume the two choices are independent." Maybe this is another thing I'm getting wrong. At any rate, I figure the answer is: probability of drawing a sock of one color the first time * probability of drawing a sock of the same color the second time = 1/3 * 1/5 = 1/15 But that doesn't seem to be correct. Any advice or subtle hints? Thanks! Link to comment Share on other sites More sharing options...
Prometheus Posted November 19, 2017 Share Posted November 19, 2017 You've found the solution to a slightly more difficult problem: the probability of a particular colour match. But here we don't care about the particular colour, only that they match. It is a bit odd for it to mention 'uniformly' but you seem to have understood it correctly: all it means is that each sock has an equal chance to be chosen. I would have assumed that anyway - but maybe someone objected that they were different sized socks or something. I agree that choosing the second sock would be a dependent event for the problem you have answered. I think it is a subtle clue to the question you should be answering. 1 Link to comment Share on other sites More sharing options...
John Salerno Posted November 19, 2017 Author Share Posted November 19, 2017 8 hours ago, Prometheus said: You've found the solution to a slightly more difficult problem: the probability of a particular colour match. But here we don't care about the particular colour, only that they match. It is a bit odd for it to mention 'uniformly' but you seem to have understood it correctly: all it means is that each sock has an equal chance to be chosen. I would have assumed that anyway - but maybe someone objected that they were different sized socks or something. I agree that choosing the second sock would be a dependent event for the problem you have answered. I think it is a subtle clue to the question you should be answering. Thank you! After thinking about your comments for a few minutes, I finally figured it out! It was mainly your first sentence that led me to the answer! But I'm still a little confused about why it says to assume both events are independent. Isn't the second choice necessarily dependent on the first event? Link to comment Share on other sites More sharing options...
Prometheus Posted November 19, 2017 Share Posted November 19, 2017 As we don't care about the colour of the first sock there is really only one event here (in a probabilistic sense). We could consider the equivalent problem. We already have one sock. There are 5 socks in a draw, one of the same colour you already have , two of some other colour and two of yet another colour. What's the probability of drawing a match for your sock? Put like this we can see there's only one event. So strictly speaking i'd say they are incorrect to describe these as independent events as there is just the one. I suspect they put it in there to help prod people away from thinking that they need to consider the probability of drawing the first sock. Or maybe just an honest mistake. 1 Link to comment Share on other sites More sharing options...
Strange Posted November 19, 2017 Share Posted November 19, 2017 52 minutes ago, Prometheus said: As we don't care about the colour of the first sock there is really only one event here (in a probabilistic sense). I think there are two events: picking the first sock and picking the second sock. They are independent in that the first has no effect on the second (other than changing the number of socks available). And, this is important because it means you can simply multiply the probabilities to get the answer. If by "only one event" you mean that you can forget about the first sock chosen when it comes to the picking second one, then that is what independent means. Link to comment Share on other sites More sharing options...
DrKrettin Posted November 19, 2017 Share Posted November 19, 2017 8 minutes ago, Strange said: I think there are two events: picking the first sock and picking the second sock. They are independent in that the first has no effect on the second (other than changing the number of socks available). But the first does have an effect on the second in that after the first has picked a sock, the second can no longer pick that particular sock. The choice of socks for the second pick depends on the pick of the first one. I don't really understand what they mean by independent in this scenario, to me it seems redundant and confusing. Link to comment Share on other sites More sharing options...
Strange Posted November 19, 2017 Share Posted November 19, 2017 Just now, DrKrettin said: But the first does have an effect on the second in that after the first has picked a sock, the second can no longer pick that particular sock. The choice of socks for the second pick depends on the pick of the first one. I don't really understand what they mean by independent in this scenario, to me it seems redundant and confusing. The choice available has changed, but the probability of what you pick from those is still purely determined by the number of each colour remaining. So you use the same method for the first sock and the second sock, and multiply them together. Imagine instead, the socks were in a cylinder in pairs. For the first sock, the chance of it being any particular colour is purely determined by the number of different socks. The chance of the second one is no longer determined by the number of socks (it is the same colour as the first one). So you can't use the same method to calculate the probability for the first pick and the second pick. I assume that is what they mean, anyway. Maybe "independent" isn't the best word. Link to comment Share on other sites More sharing options...
Prometheus Posted November 19, 2017 Share Posted November 19, 2017 19 minutes ago, Strange said: I think there are two events: picking the first sock and picking the second sock. They are independent in that the first has no effect on the second (other than changing the number of socks available). And, this is important because it means you can simply multiply the probabilities to get the answer. If by "only one event" you mean that you can forget about the first sock chosen when it comes to the picking second one, then that is what independent means. I guess you could take the first pick to be an event with probability one. But do you agree that my problem in post 4 is equivalent to the original problem? I find it simpler to think like this for the given problem. Link to comment Share on other sites More sharing options...
Strange Posted November 19, 2017 Share Posted November 19, 2017 1 minute ago, Prometheus said: I guess you could take the first pick to be an event with probability one. But do you agree that my problem in post 4 is equivalent to the original problem? I find it simpler to think like this for the given problem. Yes, I agree. I think I was thinking of it in the same way as the OP did originally. Or maybe I was thinking about drawing out all possibilities: so if the first pick is red then ... but if the first pick is blue then ... Obviously this is a much longer route to the final answer but it sometimes helps to see what is going on and why the shortcut works. Link to comment Share on other sites More sharing options...
Prometheus Posted November 19, 2017 Share Posted November 19, 2017 True, should always learn things the proper way before finding the short cuts. Did we just demonstrate a life lesson using a probability problem? Cool. Link to comment Share on other sites More sharing options...
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