Wikipedia:Reference desk/Archives/Mathematics/2011 January 28

= January 28 =

Probability
I have two sets {A} which is {1,2,3,4,5,6,7,8,9} and {B}={1,2,3,4,5,6,7,8}. If I pick three different numbers from {A} and arrange them in descending order (i.e., if I pick 2,5,1, it would become 521) and do the same for {B}, what is the probability that the number I choose from {A} will be bigger than the number I choose from {B}? How do I solve this type of problem in general? THanks. 24.92.70.160 (talk) 02:29, 28 January 2011 (UTC)
 * I assume that you are selecting the digits independently. In this case, we can define a discrete random variable $$X$$ to be the three digit number generated from A using the procedure above, and $$Y$$ to be the three digit number generated from B using the procedure above.  If each digit has an equal chance of being selected, then we can compute the probability mass function $$p_X(k)$$, which is a function mapping real numbers $$k$$ to the probability with which $$X$$ will attain the value $$k$$.  Since there are $$_9 C_3$$ (or $$\frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1}$$ or 84) ways to select three digits without replacement, the value of $$p_X(k)$$ is $$\frac{1}{84}$$ if $$k$$ is a number that can legally be generated (being of the form $$100a + 10b + c$$ for $$a, b, c \in A$$, $$a > b > c$$, and zero otherwise.  Likewise, we can do the same for $$p_Y(k)$$, the probability that $$Y$$ will attain the value $$k$$, noting that there are 56 possible numbers, each equally likely.  The probability that the number selected from A is greater (which I will call event E) can be computed by summing over all appropriate values of the joint distribution function, $$p_{X, Y}(k_1, k_2)$$, which by independence of $$X$$ and $$Y$$ is just the product of the two pmfs we generated earlier.  So the probability our desired event occurs is $$\sum_{i \in S}\sum_{j \in T, j < i} p_{X, Y}(i, j)$$ (where S is the set of legal three digit integers with digits in descending order using digits 1 through 9, and T is the set of legal three digit integers with digits in descending order using digits 1 through 8), which is $$\sum_{i \in S}p_{X}(i)\sum_{j \in T, j < i} p_{Y}(j)$$.  This is essentially in this case $$\frac{1}{84 \cdot 56}$$ or $$\frac{1}{4704}$$ times the number of pairs of integers $$(i, j)$$ where $$i \in S$$ and $$j \in T$$.
 * I apologize if this was way too thorough. I tried to stay general as much as possible.  There are potentially some shortcuts one could take with this given problem; I tried to solve it in as general a manner as possible.  RJaguar3 &#124;  u  &#124;  t  07:05, 28 January 2011 (UTC)

With shortcuts: Either X contains the digit 9 or it doesn't. The former case happens with probability 3/9 = 1/3, and then X will necessarily start with 9 and so be larger than Y.

In the case when X does not contain a 9, then X and Y are actually identically distributed. As RJaguar computed there are 56 possibilities for each, and all equally probable. So P(X=Y | X<900) must be 1/56. By symmetry, the remaining 55/56 must split evenly between X>Y and XY | X<900) = 55/112.

Putting this together we get P(X>Y) = 1/3 · 1 + 2/3 · 55/112 = 37/56. (And I hope I haven't just done your homework for you, but the question sounded more like idle curiosity than homework). –Henning Makholm (talk) 08:10, 28 January 2011 (UTC)

pls help
can someone answer this side-question here: Reference_desk/Humanities? —Preceding unsigned comment added by 77.231.17.82 (talk) 10:29, 28 January 2011 (UTC)
 * The probability of both special cards being played in the first hand is 1/25. If the game goes to the second round, each player has four cards left, and there is 1/16 chance of both special cards being played. Make sense so far? The way we deal with "If the game goes to the second round" is via conditional probability, which is a way of addressing questions of the form "what is the probability that something happens, given the fact that some other random event has already occurred?" In this case, the chance of proceeding to round 2 is 16/25. That is the number of cases in which no special card is played: (4/5)x(4/5). So to get the probability of the two special cards being played in any round, you have to sum over each round, weighted by the probability of getting to that round. SemanticMantis (talk) 16:30, 28 January 2011 (UTC)

The Stone Cech Compactification by Ultrafilters
Hi, I have a question regarding the space $$ X = N \cup \{x\} $$ where x is a point in the remeinder of the Stone Cech Compactification of N. Soppuse that I take two countable infinite families of pairwise disjoint finit sets of X, $$ \{ f_1, f_2, f_3,...\}, \{ g_1, g_2, g_3,...\} $$ and suppose that I know that $$ \cup f_i $$ and $$ \cup g_i  $$ are disjoint sets. Does that also mean that $$ [\cup f_i] $$ and $$ [\cup g_i]  $$ are disjoint sets? ($$ [A] $$ is the closure of the set A). If this is true can somone explain me why? Thanks! Topologia clalit (talk) 14:09, 28 January 2011 (UTC)
 * Any countable set can be written as the union of countably many disjoint finite sets, so you are effectively asking whether the closures of any two disjoint infinite subsets of X are disjoint. That's almost equivalent to X being discrete, which it is not, so the property does not hold: for example, let A (as a subset of N) be any element of x (as an ultrafilter) with infinite complement, and let B be the complement of A in X (including the point x). Then A and B are disjoint, but x is in the closure of both.—Emil J. 14:37, 28 January 2011 (UTC)

Thanks! But in this case I seem to have a problem undestanding the following proof of the following claim: There exist a countable non-weakly Frechet-Urysohn space. and here is the proof: Let x be an arbitrary point of the Stone-Cech reminder ω* of the discrete space ω. Then $$ X = \omega \cup \{x\} $$ is the desired example.Indeed, it is a countable space. Let us assume that X is Weakly Frechet-Urysohn. As $$ x \in \overline\omega $$ there must br a countable infinite disjoint familly F such that x(RZ)F. Let A and B be any two infinite subfamilies of F. Then both $$ x \in \overline {\bigcup \mathbf{A}} $$ and $$ x \in \overline {\bigcup \mathbf{B}} $$ must hold. But this is impossible if we choose A and B to be disjoint subfamilly of F as in this case $$ \bigcup \mathbf{A} $$ and $$ \bigcup \mathbf{B} $$ are disjoint subsets of ω and x is an ultrafilter of ω. The thing which I don't understand is this, Why does the fact that $$ \bigcup \mathbf {A}$$ and $$ \bigcup \mathbf {B}$$ are disjoint implies that it is not possible that both $$ x \in \overline \bigcup \mathbf{A} $$ and $$ x \in \overline \bigcup \mathbf{B} $$? Topologia clalit (talk) 17:27, 30 January 2011 (UTC) Also, Here are the required definitions: Definition: A point x is called weakly Frechet Urysohn point if whenever $$ x \in \overline{A} \setminus A $$ there exists a countable infinite disjoint family F of finite subsets of A such that for every neighborhood V of x the subfamily $$ \{ F \in \mathbf{F}: F \cap V = \emptyset \} $$ is finite. If every point of a space is a weakly Frechet-Urysohn point then this space is called a Weakly Frechet Urysohn space. Definition: A point x \in X and a countable infinite disjoint family F of X are said to be in the Reznichenko relation (Rz), written x(RZ)F, if the following holds: For every neighborhod V of x, the subfamily $$ \{ F \in \mathbf{F}: F \cap V = \emptyset \} $$ is finite. Thanks for any of you who will be able to help. Topologia clalit (talk) 17:27, 30 January 2011 (UTC)

average in a class
I have a 50 in a class and I am curious as to how to find my average. I have a 75 with a weight of .4; another 75 with a weight of .4; and a 0 with a weight of .4 what do I do?Accdude92 (talk) 16:52, 28 January 2011 (UTC)


 * Because all have the same weight, you just sum them up and divide: (75+75+0)/3 = 50. If the weights were different, for example .1, .2, and .3, you would multiply by the weights and divide by the sum of the weights like (.1*75 + .2*75 + .3*0)/(.1+.2+.3). --  k a i n a w &trade; 17:22, 28 January 2011 (UTC)


 * i have another class with a grade of 70. The grades are 93;100;100;100;100(weight .25)25;25;25 (weighted .25) and a 78 with a weight of .5 I am doing $$(93*0.25)+(100*0.25)+(100*0.25)+(100*0.25)+(100*0.25)+(25*0.25)+(25*0.25)+(25*0.25)+(78*0.5)/((0.25*5)+(0.25*3)+(0.5))$$ excel says that is 72.4 not 70. What am I doing wrong?Accdude92 (talk) 17:49, 28 January 2011 (UTC)
 * You should add parentheses to make it clear that all the terms before the / are the numerator. But that expression looks right; either your average is in fact 72.4, or you have the individual grades/weights wrong, or the average is being calculated with some other algorithm (e.g., discarding the highest score).  --Tardis (talk) 19:18, 28 January 2011 (UTC)


 * The average is 72.4. You will have to ask the instructor why the average says 70 instead of 72.4. -- k a i n a w &trade; 19:23, 28 January 2011 (UTC)

Learning R
I would like to learn the statistical computer language R. Are there any free online tutorials I can use whenever I have five minutes free? I am not a computer programmer. Thanks 92.24.186.58 (talk) 20:28, 28 January 2011 (UTC)


 * The r-project has manuals that include a good introduction here. -- k a i n a w &trade; 20:36, 28 January 2011 (UTC)

Thanks, although that seems extremely dry and robotic. Are there any that are more friendly? Thanks 92.28.244.55 (talk) 21:31, 29 January 2011 (UTC)


 * Interestingly enough, this page R programming books (updated) turned up on my feed reader today. It's books and not online tutorials, but it may be of some help. -- 174.21.236.191 (talk) 19:23, 30 January 2011 (UTC)


 * Introduction to Data Technologies, by Paul Murrell is a (free, online) textbook.. Large parts of the book cover more basic concepts in the lead-up to R (check the table of contents); chapters 9, 10, and 11 are the ones you'd be most interested in. There are some references there to earlier chapters, but I think you'd mostly be okay skipping the first 8. --superioridad (discusión) 01:03, 2 February 2011 (UTC)