Six years on people are still celebrating Ed Balls Day

Ed Balls Day is the social media mishap which has been celebrated on 28 April for the past six years
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  • What the Colonizers Couldn’t Take Celebrating Dia de los Muertos with the Spirit of Resistance Dia De Los Muertos, or Day of the Dead, is a Mexican holiday celebrating dead ancestors that developed from ancient traditions from before colonization and has taken place for as long as 2,5000-3,000 years. When colonizers came to the Americas they tried to stop the indigenous Aztec people from celebrating the dead in this traditional… View On WordPress
  • Not that the man is celebrating given that he’s been on the other side now for over 37 years. But if Thelonious Monk was still among the living he’d be celebrating a century on this earth. Even though he only lived to be 64-years old, the man blazed a musical trail of legends. His work was unlike anyone before or after. He played like an alien interpreting the wonky rhythms of ragtime. His songs…View On WordPress
  • The Crevice of Balls [Photo Blog] The intentionally named crevice of balls is a very interesting place. It has golf balls all over it. Swarming with little white golf balls. Also this place is quite cheap to hit at but it’s in Shimizu-cho. It is my first time hitting golf balls towards a crevice full of balls. Cheers everyone! View On WordPress
  • im the person who has played exercise tennis for 2+ years and hits so many stray balls they accidentally hit other people’s anuses far too many times
  • i went through three years of people telling me he never touched me.i went through three years of being called an attention seeker.i went through three years having doubts in my mind that he did it because of what people said.i went through three years of people saying ‘if it did happen, you asked for it.’i went through fucking years dealing with the stuff people said. what he did fucked me up.i can’t sleep if he’s home.i can’t stay in the room he touched me in without freaking out and wanting to harm myself.i can’t make or keep friends for fear of them finding out what happened and leaving me.he ruined EVERYTHING. and i’m scared i’ll always be like this.
  • [21-11] Since people are celebrating #MugabeResignsI'm asking my self who was voting for him all these years
  • [09-10] We're celebrating 75 years of fighting for a better world. Here's to passion and people power ? #MondayMotivation
  • [30-09] We loved celebrating 30 years of #TNG30 at our @StarTrekEire event last night! Many people to thank but a shout out…
  • [11-07] So #WorldFringeDay is celebrating 70 years since the first ever fringe in #Edinburgh; it's NOT celebrating this...
  • [31-10] To win #HKSixes @OfficialCSA needs: 10 off 5 balls, 9 off 4 balls, 8 off 3 balls, 4 off 2 balls, Wicket! 4 off 1 ball, FOUR
  • [15-10] @supersteel23 Celebrating Columbus Day is celebrating genocide of many indigenous people. #ColumbusDay should be a day of mou
  • [30-09] Today, most people are celebrating the #FirstDayofFall. I'm celebrating the 41st anniversary of #CharliesAngels. :
  • [07-11] #ExplainBalls in a song. Balls are long. Balls are wrong. Plz dont take my balls with a tong! Or my cookies, or my sch
  • [08-10] #RHOC Really Lydia. You are uncomfortable dressing up as a man for a few hours, but you can say balls balls balls every episode
  • [02-11] @Shellski5 I’ve got itchy sweaty balls. You know balls. That’s just sort of a balls thing.#ExplainBalls
  • [27-07] Balls balls balls .. beats the battered mars bar #Scotland #westlothian #mealsforaquid
  • [03-10] Lydia is uncomfortable about some things, but has no problem talking about balls, balls, and more balls. #RHOC
  • [10-10] @MorningMashUp My husband and I celebrating our anniversary! 9 years ago today we met. 7 years ago today we married! #TellMeSomethingGood
  • [30-10] The #RedSox played with last years balls to save Luxury Tax space
  • [21-09] When Islamic New Years falls on int #PeaceDay?✌?️❤️ Happy #Islamic New Years to everyone celebrating! We are united in #Pe
  • [28-09] I swear..if I was in my 20-30's spent several years juggling a 86-91 yr old person's moth balls..I better be in that will #RIPHefner
  • [03-11] So confused as to why the #Jets hate Bilal Powell so much now. The guy catches a ton of balls the last 2 years and now no
  • [14-11] @jasonmatheson This was probably day after Ramadan, families celebrating. I've seen that. Plus who cares! People are people. #stopthehate
  • [21-11] "Black, white... rich people, poor people, we're celebrating" - former farmer and activist Ben Freeth #MugabeResigns :
  • [06-09] Lots of people are celebrating the return of #GBBO by holding a #BigTea party to help older people! Find out more:
  • [03-10] They should've had a balls voyage party 4 David Beador years ago, cause his wife emasculates him daily #RHOC
  • [01-10] Celebrating 35 years! #Epcot35
  • [12-11] Celebrating 21 years as Digital Experts
  • [17-09] Celebrating 23 years of friendship! #bostoncollege…
  • [21-11] Since people are celebrating #MugabeResignsI'm asking my self who was voting for him all these years
  • [14-11] @jasonmatheson This was probably day after Ramadan, families celebrating. I've seen that. Plus who cares! People are people. #stopthehate
Six years on people are still celebrating Ed Balls Day
Ed Balls Day is the social media mishap which has been celebrated on 28 April for the past six years
How people lived before 10-12 years when there was no internet ?
Should people be advising on STEM, as if it's going to take 10-20 years to get that education and experience, it may not be in demand then?
Lastly, should the responsibility of celebrating the Rohingyas be borne alone in Bangladesh?
Are there more male Surgeons because men got years of experience by shaving their balls?
Celebrating 50 Years of '100 Years of Solitude'
First published in 1967, Gabriel Garcia Marquez’s modern classic has touched millions of readers, including former U.S. presidents Bill Clinton and Barack Obama.
ELI5: Why some people get the "heebie jeebies" when they touch certain items? (ie sponge foam, peaches, cotton balls, etc)
Celebrating 80 years
Celebrating 50 years
Celebrating 40 years of WTA!
Celebrating 25 years
Celebrating 50 years of NSD
Celebrating 60 years of INP
Celebrating 60 years
Celebrating 125 years
Two people draw 3 balls each out of a bag of 13 balls. Various conditions apply. What are the outcomes? A bag contains 13 balls. Person $A$ chooses randomly 3 balls out of the bag, afterwards Person $B$ does the same. Let $E_A$ be the event where $A$ chooses 3 balls of the same colour. Let $E_B$ be the event where $B$ chooses 3 balls of the same colour. Let $F_A$ be the event where $A$ chooses exactly 2 balls of the same colour. Let $F_B$ be the event where $B$ chooses exactly 2 balls of the same colour. Let $G$ be the event where $A$ and $B$ have exactly the same selection. Consider: a) the balls have different colours; b) 5 balls are blue, 5 balls are red and 3 balls are green How many different outcome are possible if: i) $A$ places the balls back before $B$ chooses. ii) $A$ does not replace the balls. $E_A$ $E_AF_B$ $(E_A \bigcup E_B)^c(F_A \bigcup F_B)^c$ $F_AF_BG$ $F_A \bigcup G^c$ $E_A E_B G$ My problem: I have an old memo but I don't agree with the solutions. Their solution for: (a.i.5) $| F_A \bigcup G^c | = {13 \choose 3}{13 \choose 3} = 81796 $ My problem: If $G^c$ means $A$ and $B$ have different selections then surely we count ${13 \choose 3}$ too much as somewhere $B$ will have the same ordering as $A$. As such we have to subtract this amount from the above solution they got? (b.i.1) $ | E_A | = [{5 \choose 3} +{5 \choose 3} + {3 \choose 3}]{13 \choose 3}$ My problem: I don't get the multiplication with ${13 \choose 3}$ b.i.2. and b.ii.2. their solution respectively $| E_A F_B | = [ {5 \choose 3} + {5 \choose 3 } +{3 \choose 3}][{5 \choose 2}{8 \choose 1}+{5 \choose 2}{8 \choose 1} + {3 \choose 2}{10 \choose 1}]$ $| E_AF_B | = {5 \choose 3}[ {2 \choose 2} {8 \choose 1 } + {5 \choose 2} {5 \choose 1} + {3 \choose 2} {7 \choose 1}] + {5 \choose 3}[ {5 \choose 2} {5 \choose 1} + {2 \choose 2} {8 \choose 1 } + {3 \choose 2} {7 \choose 1}] + {3 \choose 3}[ {5 \choose 2} {5 \choose 1} + {5 \choose 2}{5 \choose 1} ] $ My problem: Does $E_A F_B $ not form disjointed sets and therefore $| E_A F_B |$ = 0 for both the above cases? ++++++++++ ADDITION 1 ++++++++++ Why is $|E_A| = ({5 \choose 3} +{5 \choose 3} + {3 \choose 3} ){13 \choose 3} $ Rather than just $|E_A| = {5 \choose 3} +{5 \choose 3} + {3 \choose 3} ? $ So it seems my problem is in understanding exactly how to interpret an event. In my mind $|E_A|$ looks only at event E for A irrespective of what B draws, because if I want to calculate the probability of person A drawing only 3 balls of the same colour then surely the probability is independent of whatever balls B may draw?? I still don't get this. The only way that I can somehow make sense of it is if we look at the solutions of the sample space, the elements of which is a solution set for all A and B selections, as the question states that the 2 events occur (have to and always will) one after the other. My issue still is that working out the probability of $E_A$ will erroneously take into consideration the selections for B. Ok, so it just occurred to me that in fact if we do work out the probability of $E_A$ we can do it in two ways 1) we only look at the draw of A and therefore work with a reduced sample space; and 2) when taking the entire sample space into consideration, thus: For (1) we have: $ P(E_A) = \frac{| E_A |}{|S|} $ … where for S and thus E we look at the reduced sample space i.e. we don't take B into consideration. $ = \frac{{5 \choose 3} +{5 \choose 3} + {3 \choose 3}}{ {13 \choose 3} }$ $ = \frac{21}{286}$ And for (2) $ P(E_A) = \frac{| E_A |}{|S|} $… where for S and thus E we look at the original sample space $ = \frac{ [ {5 \choose 3} +{5 \choose 3} + {3 \choose 3}] [{13 \choose 3}] } { {13 \choose 3} {13 \choose 3} }$ $= \frac{21}{286} $ So this seems to indicate once again that the only reason we take into consideration the draw of B is because the defined solution space enforces it?? I would think that there would be some more elegant way to express this in the Math as it is not evident when you just look at a variable defined as $ E_A $? Surely I am missing something? Let me try explaining this way: Why is $|E_A| = [{5 \choose 3} +{5 \choose 3} + {3 \choose 3}] {13 \choose 3} $ Rather than just $|E_A| = {5 \choose 3} +{5 \choose 3} + {3 \choose 3} ? $ Consider for a moment that we have 5 Blue, 5 Red and 3 Green balls. I am trying to find the flaw in the following argument: For $(E_A \bigcup E_B)^c (F_A \bigcup F_B)^c $ consider first $(E_A \bigcup E_B)^c = {E_A}^c \bigcap {E_B}^c$ Let's look at the elements in $ {E_A}^c $: $E_A(B, R, G) = { (1,1,1); (2,1,0); (2,0,1); (1,2,0); (0,2,1); (1,0,2); (0,1,2) }$ … And all permutations of the colours in every element But ${E_B}^c$ has exactly the same set i.e. $E_B(B, R, G) = { (1,1,1); (2,1,0); (2,0,1); (1,2,0); (0,2,1); (1,0,2); (0,1,2) }$ So that ${E_A}^c \bigcap {E_B}^c = { (1,1,1); (2,1,0); (2,0,1); (1,2,0); (0,2,1); (1,0,2); (0,1,2) }$ according to the enforcement of the AND operator. Now for $(F_A \bigcup F_B)^c = {F_A}^c \bigcap {F_B}^c$ Let's look at the elements in $ {F_A}^c $: ${F_A}^c = { (1,1,1); (3,0,0); (0,3,0); (0,0,3) } $ also ${F_B}^c = { (1,1,1); (3,0,0); (0,3,0); (0,0,3) }$ So that ${F_A}^c \bigcap {F_B}^ = { (1,1,1); (3,0,0); (0,3,0); (0,0,3) }$ …. And all permutations of the colours in every element. Taking the above it follows that $(E_A \bigcup E_B)^c (F_A \bigcup F_B)^c = ( { {E_A}^c \bigcap {E_B}^c } ) \bigcap ( { {F_A}^c \bigcap {F_B}^c } )$ $= { (1,1,1) }$ … as this is the only element that is in their intersection. It should therefore follow that: $ | (E_A \bigcup E_B)^c (F_A \bigcup F_B)^c | = {5 \choose 1} {5 \choose 1} {3 \choose 1} $ BUT the solution in the memo is: $ {5 \choose 1} {5 \choose 1} {3 \choose 1} . {5 \choose 1} {5 \choose 1} {3 \choose 1} $ If I can figure this out then most of my problems should be solved ... I hope ...
Did Puritans punish people for celebrating Christmas? The Facebook group Americans for the Separation of Church and State posted this link: Puritans in Massachusetts arrested and fined anyone who celebrated Christmas. But go ahead, tell us how awful ...
Are people happier now than 100 years ago? 1000 years ago? 10000 years ago? [closed] I don't know if this goes under philosophy, but I made an account here because I got really curious about this. I feel like the more materialistic society gets, we get less and less happy, but I am ...
A box contains 100 balls of different colours.Smallest no of balls drawn from box so that at least 15 balls are of same colour is A box contains 100 balls of different colours: 28 red 17 blue 21 green 10 white 12 yellow and 12 black.Smallest no of balls drawn from box so that at least 15 balls are of same colour is Well , I had been trying to solve this question the other way round. I am assuming first that I have taken largest no. of ball that is 100 from the box. Then I had gradually been decreasing the number of balls I had taken out from 100 to 99. Like if I keep one ball back in the box then the probability of at least 15 balls out of the box would be hundred percent. (That is at least 15 balls of same colour must be out of the box). Extending this approach I think the answer should be 100-(28+21+2) [all red, all green and 2 blue still in box.] But the answer comes out to be 77. I am not able to understand what went wrong
What is the probability of selecting 2 balls from a bag of $n$ blue balls and $m$ red balls, if... Say there are $n$ blue balls and $m$ red balls in a bag. What is the probability of selecting 2 balls if exactly 1 is red and the other is blue? Note: If these balls are selected at the same time (i.e. I put my hand in the bag and pull out two balls simultaneous), will the probability be the same as if I selected them separately? (i.e. I put my hand in the bag and pull out one, then put my hand in again and pull out the second one.) From trying with a few values of $n$ and $m$, it appears the probability is the same in both cases, which is a bit surprising. $ $ Any clarification would be great.
Did people use to urinate everywhere at fancy balls? My history teacher once told us about how several centuries ago people in Europe who were guests at fancy balls and masquerades used to urinate on curtains, under stairs and in dark corners of dimly ...
Why do basketballs bounce higher than ping pong balls tennis balls styro foam balls and golf balls? Hmmm ok .. lets get to physics.. its basically due to energy content... when you hold a ball it has some sort of energy due to gravitation called the potential energy .. nd that potential energy is related to the mass.... now whn u throw the ball down.. all (can be less) the potential energy gets converted into the kinetic energy after collision .. nd so... the basket ball having higher mass... have high pot. energy nd thus high KE aftr collision nd thus high velocity (speed) nd thus high altitude .. So one word answer : Energy
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Solution to a bag contains 4 balls Two balls are drawn at random and are found to be white What is the probability that all balls are white? Two balls are certainly white. So the remaining balls may be one of the following{WW, WN, NW, NN}where W means a white ball and N means non-white ballLet the events be:E- all four balls are white(WW)F- 3 balls are white(WN, NW)G- only two balls are white(NN)Now, P(E) = P(G) = 1/4, and P(F) = 1/2Let A be the event that the two balls drawn are whiteP(A|E) = 4C2/4C2 = 1P(A|F) = 3C2/4C2 = 1/2P(A|G) = 2C2/4C2 = 1/6Now By Bayes' Theorem,required probability = P(E|A) = (1*1/4) / ( 1*1/4 + 1/2*1/2 + 1/6*1/4 ) = (1/4) / ( 13/24 ) = 6/13
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