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About Tullius

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  1. No, that is not why quantum computers are interesting. Instead, they allow for some really fast algorithms to be used. One example is this one: On a classical computer, searching for a single element in a unsorted list of N elements takes O(N) time, i.e. the time taken is approximately proportional to the length of the list. If the list is sorted, this can be improved to O(log(N)), i.e. the time taken is approximately proportional to the logarithm of the length of the list. On a quantum computer, something weird happens for this problem: There exists an algorithm that can search for an element in an unsorted list and find it with high probability in just O(log(N)) time, i.e. just as fast as for the sorted list on a classical computer. So, on a classical computer, searching for one element in an unsorted list with 1 trillion elements will take a million times longer than for the unsorted list with 1 million elements. On the other hand, on a quantum computer, it would only take twice as long. What you are describing is still a classical computer and so, even if your computer is capable of finding one element within an unsorted list with googol (10^100) elements within 1 day, it won't be able to do so for a list with googol^2 elements before the end of the universe (as it would take you googol days). If a quantum computer could do this for the case of googol elements within 1 day, it would only take 2 days for googol^2 elements.
  2. I think it has more to do with the lack of planned Soyuz launches (according to https://en.wikipedia.org/wiki/List_of_Russian_manned_space_missions): There are 3 launches scheduled for 2019 and only 2 scheduled for 2020, instead of the 4 we have been used to since 2010. So if they want to maintain 6 astronauts on the ISS with rotations every 6 months, they either need more Soyuz's, or Crew Dragon or CST-100 need to fly. And if they want to do more Soyuz launches, because Crew Dragon and CST-100 are not yet flying, there has to be a treaty between NASA and Roskosmos to decide who pays for them and Congress needs to allocate additional funds for it, besides the obvious problem of building more Soyuz's on short notice. And we all know how much time such an endeavour would take. So the contingency plans will probably revolve around prolonging crew stays on the ISS or reducing the crew complement of the ISS.
  3. I think some people here fail to see the sheer size of transporting an atmosphere to Mars. Relevant xkcd-What-if: https://what-if.xkcd.com/153/ Since Mars has only half the radius of Earth, the pile would only need to be a quarter of the size it has for Earth. However, since Mars has lower gravity, if we want the same surface pressure, we need a higher atmosphere.
  4. Or they just built a Block 5B booster as the Block 5 variant for the FH core.
  5. Let us use coins instead of kids. For two coin tosses, we have the following possible outcomes: Each of the possible outcomes is of course equally likely, i.e. any single one of the cases HH, HT, TH and TT has probability 1/4. So the likelihood of obtaining exactly 1 heads is 1/2. And you cannot change the probabilities by mixing the coins after the toss. (The coins are flipped, so you won't change the result by simply moving the coins around) So according to the diagram above, I flip two coins. Now, I tell you that at least one of the coins is showing heads, i.e. number of heads >= 1. And nothing more, I neither show you the coin with heads, nor do I tell you if it was the first toss or the second toss. What is the probability that the two coins are showing heads?
  6. You could call them kid A and kid B based on the order they came in through the door yesterday or based on the grades of their last math tests. It doesn't really matter. You could even take one dollar coin and one Euro coin, if you want to think of coin flipping. If there is exactly one heads after flipping both, either the dollar is heads and the Euro is tails, or the dollar is tails and the Euro is heads. And, of course these two possibilities are very distinct, although both amount to one heads and one tails. And coins or kids, doesn't make a difference. Another way you may look at it: You could just compute that P (two girls) = P (kid A girl) * P (kid B girl) = (1/2) * (1/2) = 1/4 P (two boys) = P (kid A boy) * P (kid B boy) = (1/2) * (1/2) = 1/4 using that the gender of both kids is independent. So you have P (same gender) = P (two girls) + P (two boys) = 1/4 + 1/4 = 1/2 Since, probabilities always have to add up to 1, we therefore must have P (different gender) = 1 - P (same gender) = 1 - 1/2 = 1/2 So, it is twice as likely that the kids have a different gender (i.e. one boy and one girl) than the kids being both boys.
  7. First, even for twins, one is usually older than the other by a few minutes. However, you could just replace oldest and youngest by any other choice of identifiers (e.g. genderneutral names), But, let's just start with a much simpler problem and just look at the different possibilities parents can have two children: case oldest kid youngest kid 1 girl girl 2 girl boy 3 boy girl 4 boy boy And we can hopefully agree that each of these cases is distinct and in our mathematical context equally likely, i.e. each one has probability 1/4. So what is the probability of two girls? 1 case among 4, and so the probability is 1/4. What is the probability of one boy and one girl? 2 cases among 4, and so the probability is 2/4 = 1/2. And now the million dollar question: In how many cases of these four cases will you say that there is at least one boy?
  8. I would rather say, if we use A and B as the identifiers of the kids (and you need some form of identifiers): A is boy, B is girl A is girl, B is boy A is boy, B is boy These are the three possible cases in which you will say that there is at least one boy.
  9. It is NOT independent! You are not telling me that the oldest kid is a boy, but that at least one of the kids is a boy. It could be either the oldest or the youngest, and I DON'T know. And that is the important bit, as you say that at least one of the kids is a boy, if either the oldest or the youngest or both is a boy. And so there are THREE cases of equal probability, in which you say that there is at least one boy. And there is only ONE case, in which we have two boys (i.e. the second one is a boy too). Therefore the probability is 1/3. And, if you still don't believe me, just look at the calculation I posted in my previous post.
  10. But which one is hand one or hand two? If you keep both hands behind your back at all times, I don't know in which one the first red crayon is. It can either be the right hand or the left hand. And then the question is, if there is a red crayon in the second hand. In that case, it will be 1/3. If you show me your first hand with the red crayon, there are only two possibilities left, as I know which one is the first hand. And, if graphs confuse you, maybe some mathematical computations don't:
  11. Cases 1 and 2 are not the same. Let's imagine you flip a coin to decide if you put a red (heads) or orange (tails) crayon into a box (a process which I can't see). And you do this twice, as we want two crayons in the box. Then you shake the box, so it is impossible to know which crayon was put in first. You look into it and say to me: "There is at least 1 red crayon in the box". Now, I know that either on the first flip, on the second flip or on both flips, you got a heads, which means that there are 3 different possible cases and only in 1 we end up with two red crayons. If instead you told me that the first crayon you put in the box was red (or the crayon in your right hand is red), then of course the probability of two red crayons would be 1/2, as there only two possibilities: the first is red and the second is red the first is red and the second is orange
  12. Well, if you have at least one red crayon in your hands, we have the following possible situations: case right hand left hand 1 red orange 2 orange red 3 red red So the probability of two red crayons is 1/3.
  13. Cases 2 and 3 are NOT equal. They are the two possibilities to get 1 boy and 1 girl. For you example with the crayons: You have one crayon in each of your hands and they can be red or orange. The possible cases are of course: case right hand left hand 1 orange orange 2 red orange 3 orange red 4 red red There are three cases in which you will say that you have at least one red crayon (cases 2, 3 and 4), but only in one of them (case 4) the two crayons will be red, i.e. if the probability of each case is equal, we have P (2 red crayons | at least 1 red crayon) = 1/3
  14. If we agree that "one of them is a boy" means "at least one of them is a boy", we have the following: case kid A kid B 1 girl girl 2 boy girl 3 girl boy 4 boy boy So there are three cases, where the parent will say "one of them is a boy", but only one case, where the second kid is also a boy. As we assume for simplicity that all cases are equally likely, the answer is 1/3. The reason is that the boy is not identified, i.e. you don't know in which of the two prams the boy is (nor do you know anything else identifying the boy) If the parent would point at a pram and say "this one contains a boy" (or he would just say "Joey is a boy"), the kid is identified (i.e. in terms of the table above, we have "kid A is a boy") and so the correct answer is 1/2.
  15. If you only teach the concept using the standard set mathematical wording, you don't teach them something about practically using it (Very rarely in practice, you will find the precise wording "at least x of y"). You want students to stumble over the problem, when they try to intuitively give an answer to the question, so that they learn that they should only trust their mathematical computations. Intuitively, you would say that if the mother would tell you that "one of them is a boy" or "Tony is a boy" doesn't make any difference. However, this makes all of the difference, cf. my previous post. It is the same with your example: Unless you tripped over this problem, you will make this error time and time again.