A disk 2 inches in diameter is thrown at random on a tiled floor, where each tile is a square with sides 4 inches in length. Let C be the event that the disk will land entirely on one tile. In order to assign a value to P(C), consider the center of the disk. In what region must the center lie to ensure that the disk lies entirely on one tile? If you draw a picture, it should be clear that the center must lie within a square having sides of length 2 and with its center coincident with the center of a tile. Since the area of this square is 4 and the area of a tile is 16, it makes sense to let $P(C) = \frac{4}{16}.$  I don't really understand the statement for the last sentence "Since the area of this square is 4 and the area of a tile is 16, it makes sense to let $P(C) = \frac{4}{16}.$"
121760
I came up with a mental problem in probability and I found out that it's either trivial or more complex than what I thought. The problem is this:  Take a coin of diameter $D$ and radius $R$, and toss it. Cover the table with a rectangular tablecloth made of small squares of side $\ell$, with $\ell \geq D$. What is the probability for the coin to land exactly inside a little square?  If the problem has a trivial solution then I expect it to be this: the probability is the area of the coin over the area of the square: $$P = \frac{\pi R^2}{\ell^2}$$ But here I would a sort of $\theta$ function such that (in case this hypothesis is missing) the probability is zero when $D&gt;\ell$ hence: $$P = \frac{\pi R^2}{\ell^2}\theta(\ell - 2R)$$ But here many problems arise: first of all my convention demands that $\theta (\ell - 2R) = 1$ for $2R = \ell$, because as improbable it is, there could be a golden toss by which the coin lands exactly perfectly within the square the side of which is identical to the diameter. But now I strongly believe I shall consider the whole number of squares of the tablecloth. But more squares = more probability? I'm stuck on an either trivial or complicated reasoning. Any clarification?
108804
So basically the question is to find the CDF of $Z$ where $Z$ is the random variable that signifies the distance from a point in a square(sides 1 length) to a fixed vertex of the square. I do not understand the calculation of the geometry in the following picture :
38535
Say there are $p_{1}$ red balls and $p_{2}$ green balls. We put all the balls in a circle with $p_{1}+p_{2}$ places in total. It is forbidden that a ball (red or green) is placed between two red balls.    Show that this is impossible if $p_{1}&gt;p_{2}$.   Further on, say $p_{1}=p_{2}=p\geq10$. For which values of $p$ the ordination is possible?  Can somebody help me? Thanx in advance!
83982
I've seen many variations of this problem but I can't find a good, thorough explanation on how to solve it. I'm not just looking for a solution, but a step-by-step explanation on how to derive the solution.  So the problem at hand is:   You have m balls and n bins. Consider throwing each ball into a bin uniformly and at random.    What is the expected number of bins that are empty, in terms of m and n? What is the expected number of bins that contain exactly 1 ball, in terms of m and n?   How would I approach solving this problem?  Thanks!
11856
I get the graphic explanation, i.e. that the area $dA$ of the sector's increment can be looked upon as a polar "rectangle" as $dr$ and $d\theta$ are infinitesimal, but how do you prove this rigorously?    1. Geometrically, the exact area would be $$\frac{(r+dr)^2d\theta}{2} - \frac{r^2d\theta}{2}$$$$= (r + \frac{dr}{2}) dr d\theta  $$$$= r dr d\theta + \frac{dr^2 d\theta}{2}.$$  How do we get rid of $\frac{dr^2 d\theta}{2}$? Is it too insignificant in value compared to $r dr d\theta$ so that it vanishes? If so, is it the reason it gets ignored? And what's more, geometrically speaking, why would the two areas - $dx dy$ and $r dr d\theta$, be equal?  2. Approaching it algebraically, setting $$x = r\sin\theta$$$$y = r \cos\theta$$  gives $$\frac{dx}{d\theta} = r\cos\theta,  dx = r\cos{\theta} d\theta$$$$ \frac{dy}{dr} = \cos\theta,  dy = \cos\theta dr. $$  Multiplying both equations, side by side, gives $$dxdy = r\cos^2\theta dr d\theta.$$  Again I get an extra term, which is $\cos^2 \theta$. In both cases I am unable to derive that $dxdy = rdrd\theta$.  What do I do wrong in my reasoning? It all makes me think that I'm getting something essential terribly wrong.
24075
I am preparing a presentation on statistics. I want to include a slide on Bayes' Theorem which says:  p(\theta | D) = \frac{p(D|\theta) p(\theta)} {p(D)}   I want to explain each term using arrows. How can I do this?  I tried to follow this  but my main problem is that I have a fraction. How can I split the fraction?   is exactly what I want to do. Without the last line.
65067
I am having difficulties understanding and solving the following problem.    So far I think I understand the first line. I know what the graph of floor[t] looks like and I understand they're asking to integrate from 0 to some value of n and that integral always equals the equation on the right. I do not really understand what they are asking for in the second line or how that differs from the first equation.  I have checked through my notes and various online sources and we've just covered the fundamental theorem of calculus and discussed integration using partitions on an interval to calculate area but I am also unsure about how to relate these to this problem if they are even needed.  Any pointers, explanations or solutions would be greatly appreciated.  Thank you.
96887
I'm trying to figure out what is the probability with 3 D10's if I were to roll them one after another."001" is one and "000" is a thousand. Would every number have the same probability or doubles and triples be less common?  And please explain.
98724
It is claimed that probability of continuous R.V. is 0 for ALL x in R. That is, for every point the probability is zero.  But, somehow when we sum all these zero probabilities over the entire domain $\mathbb{R}$, the total becomes equal to 1.    Can somebody explain how 0s can add up to 1?  This defies kindergarden math.
66158
First, I apologise for the vague title. I couldn't think of a short way to represent the problem. Also, I am aware that a similar question exists, but I have a little bit more insight.  The problem is:      A man is known to speak the truth three out of four times. He throws a   die and reports that it is a six. Find the probability that it   actually is a six.   The answer in the book is $ \frac{3}{8} $, which sounds completely off to me, given the numbers, on first glance. I decided to try it on my own. After some contemplation, it indeed is true that $ \frac{3}{8} $ is correct. But, the question's vagueness makes it interesting. It's sort of a naive answer!  First, here's the book's solution:  Let $ E $ be the event that the man reports that a six occurs, $ S_1 $ be the event that six occurs and $ S_2 $ be the event that six does not occur.  $ P(S_1) $ = Probability that six occurs = $ \frac{1}{6} $  $ P(S_2) $ = Probability that six does not occur = $ \frac{5}{6} $  $ P(E~|S_1) $ = Probability that the man reports that a six occurred when it has actually occurred (truth) = $ \frac{3}{4} $  $ P(E~|S_2) $ = Probability that the man reports that a six occurred when it hasn't actually occurred = $ \frac{1}{4} $  $ P(S_1|E) $ = Probability that a six has actually occurred when the man claims so =  $$ \frac{P(S_1)~P(E~|S1)}{P(S_1)~P(E~|S_1)+P(S_2)~P(E~|S_2)} = \frac{3}{8} $$  Link to original: Page 25 of  (link updated 22/11/2015; tends to break often).  Here's the first of my two solutions:    The question is a little vague and it doesn't tell us what the man says when he doesn't get a six. Assuming he lies every time he doesn't get a six (which is not quite what the question says), would bring up $ P(E~|S_2) $ to $ 1 $, which skyrockets $ P(lie) $ to $ \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6}.\frac{1}{4} = \frac{7}{8} $.  Also, $ P(E~|S_1) = \frac{3}{4} $ and $ P(E) = \frac{5}{6} + \frac{1}{6}.\frac{3}{4} = \frac{23}{24} $.  This brings us to, $$ P(S_1|E) = \frac{P(S_1)P(E~|S_1)}{P(E)} = \frac{3}{23} = 13.04\% $$  Please note that I don't care about the sixes rolled when he lies about it being a six, because that condition is not applicable to the question.  I'm very skeptical about my calculations! I decided to whip up a simple program and see what the probability distribution is for $$ \frac{Number~of~sixes~rolled~while ~saying~it's~a~six}{Number~of~dice~rolls} $$  Here's the result:    Considering the book says $ P(E~|S_2) = \frac{1}{4} $, which is in essence is contradictory to the question, let's also consider the other possibility:    It's even more vague in this scenario! If the man lies after any given roll, there is a $ \frac{1}{5} $ probability that he will lie about it being any other number. This creates a new branch at each lie, which splits into five. In this case,  $$ P(E) = \frac{1}{6}.\frac{1}{4}.\frac{1}{5}.\frac{5}{1}+\frac{1}{6}.\frac{3}{4} =  \frac{1}{6} $$  $ P(E~|S_1) $ still remains $ \frac{3}{4} $, but $ P(E~|S_2) = \frac{1}{4}.\frac{1}{5} = \frac{1}{20} $.  $$ P(S_1|E) = \frac{P(S_1)P(E~|S_1)}{P(E)} = \frac{3}{4} = 75\% $$  This seems like the more intuitive answer, and it is why I doubted the original answer in the first place. Given a truth probability of $ \frac{3}{4} $, there was no way that the probability of a six was $ 37.5\% $!    The ambiguity of the question is what makes $ \frac{3}{8} $ wrong to me. Entirely out of curiosity, which answer fits best with the question and are these calculations correct or baloney?
470
On the , there is this line:      $p(\mathrm{height}|\mathrm{male}) = 1.5789$ (A probability distribution over 1 is OK. It is the area under the bell curve that is equal to 1.)    How can a value $&gt;1$ be OK? I thought all probability values were expressed in the range $0 \leq p \leq 1$. Furthermore, given that it is possible to have such a value, how is that value obtained in the example shown on the page?
107193
The dice are fair.  You have a $1\over6$ chance of getting the first number. A $1\over6$ chance of the second and so on.  Is it just $({1\over6})^3$ (1/216) or is that not accounting for the second and third roll properly?
71202
My reasoning is as follows: the p.d.f. is divided by the mean (expected value) into two parts, for which the areas under the p.d.f. curve are equal, hence the probabilities that random variable takes a value less then or equal to the mean are 0.5, which means that that value is also the median (by definition of median).  But textbooks say that the mean and median are equal only if the p.d.f. is symmetrical. Where is my mistake?
84249
A coin is tossed three times. The probability of zero heads is 1/8 and the probability of zero tails is 1/8. What is the probability that there is at least one head and at least one tail?  So, if P(zero heads)= 1/8 , then that should be the same of p(all tails)?  We would use the complement rule and Multiplication Rule?   P(at least one head) = 1 - P(no heads) = 1 - 1/8= 7/8 P(at least one tail) = 1 - P(no tails) = 1 - 1/8= 7/8  would I multiply the two values?
116612
I have the following sentence:  If I toss a biased coin with 2/3 chance of landing on heads, given that there was at least one head in 3 flips, what is the probability that there is only 1 head?  If we let $X$ represent the number of heads, then I can use $Baye's$ formula to find out the probability.     Which of the following would be the formula?  $$P(X=1 | X \geq 1)$$ $$P(X \geq 1 | X=1)$$  I think its the first one, but my sheet is saying the second one. I think it might have made a mistake. Which is the right probability?
19798
Are the floor functions of $0.999\cdots$ and 1 equal?  It is true that $0.999\cdots=1$ but how does one justifies the integer part of $0.999\cdots$ being 1 , where it is not, or alternatively without using $0.999\cdots=1$ how can we show that  $\lfloor0.999\cdots\rfloor = \lfloor 1 \rfloor$ ?
70706
The present question follows on from some other questions on this site asking how to generate uniform points inside a disc (see e.g., ,  and ).  The natural extension of that problem is to generate points inside an $m$-dimensional   with centre $\mathbf{c} \in \mathbb{R}^m$ and radius $r \geqslant 0$.  That is, we want to generate IID random variables from the following distribution: $$\mathbf{X} \sim \text{U}(\mathcal{B}(\mathbf{c},r)) \quad \quad \quad  \mathcal{B}(\mathbf{c},r) \equiv \Big\{ \mathbf{x} \in \mathbb{R}^m \Big| ||\mathbf{x} - \mathbf{c}|| \leqslant r \Big\}.$$ How do we generate IID uniform points on this space?  Is there a simple way to program this?