Ed Balls Day is the social media mishap which has been celebrated on 28 April for the past six years
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
How many years have you been married if you are celebrating your diamond anniversary?
When did people first start celebrating Hanukkah?
When did people start celebrating Christmas?
What do people eat when they are celebrating saint David's day?
The traditional St.Davids Day meal starts off with cawl (pronounced 'cowl'), a hearty broth of mutton & vegetables, and is followed by leeks with either bacon or pork in cheese sauce. Lavabread (made from lava seaweed) is sometimes eaten on toast as a side dish.Variations on the St.David's Day feast include Welsh lamb served with leeks, potatoes and gravy, and starters of either Welsh rarebit (a spicey cheese sauce on toast containing mustard, honey and onions) or lavabread patties.
What age should people stop celebrating Halloween?
When do people in Mexico start celebrating Christmas?
Why dont equal masses of golf balls and ping pong balls contain the same number of balls?
Frances hits a cue ball into a rack of billiard balls sending the balls in many different directions. After a few seconds all the balls on the table stop moving. Which of the following is true?
How much were people fined in Boston in the 1600s for celebrating Christmas?
Which country has the lowest amount of people celebrating Christmas?
Will topflight golf balls ever make yellow balls for their freak brand of balls?
yes
People in the us make a bigger deal of celebrating st Patrick's day than those in Ireland?
What is the feast celebrating the andel of death sparing the Jewish people?
Do people in the us make a bigger deal of celebrating Stpatrick's day than those in Ireland?
How do you put ten identical balls on a surface so that you can have 5 lines of balls with 4 balls on each line?
Why is there no mention in the Bible of Christians celebrating Jesus birthday even though the last book of the Bible was written more than 60 years after he died?
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