I read in multiple places a math-related thing saying that if there are 30 people in a room, the odds aren't 30/365 that they will have the same birthday, but more like a 90% chance. Here is an explanaition of it: http://www.mste.uiuc.edu/java/java/birthday/explanation.html\n\nHowever, it does not make sense to me. Can anybody explain it out?
215997
Here's a step by step explanation of how the probability is assessed.  Hope this will make things clearer for you.\n\nhttp://www.mste.uiuc.edu/java/java/birthday/explanation2.html
103962
In the experiment of choosing the 3 digits\n\nTotal no. of possible choices \n= (no.of ways in which the first digit can be chosen) * (no.of ways in which the second digit can be chosen) * (no.of ways in which the third digit can be chosen)\n==> n = 9 * 10 * 10\n==> n = 900\n\n[Since the first digit cannot be zero, there would be 9 possibilities for choosing it, 10 for second and third]\n\nLet "A" be the event that all the 3 digits are different and "B' the event that the 3rd digit is different from the first 2 digits.\n\nFor Event "A"\n\nNo.of favorable choices\n= (no.of ways in which the first digit can be chosen) * (no.of ways in which the second digit can be chosen) * (no.of ways in which the third digit can be chosen)\n= 9 * 9 * 8\n==> m(a) = 648\n\n[Since the three digits to be chosen should be different, there would be 9 possibilities for choosing the first digit (since 0 is excluded), 9 possibilities for the second (excluding the one chosen for the first and including 0) and 8 for the third (excluding the ones chosen for the first and the second and including 0)\n\nProbability that all the digits are different\n==> Probabilty of occurance of event "A" = m(a)/n\n==> P(A) = 648/900\n==> P(A) = 18/25\n\nFor Event "B"\n\nNo.of favorable choices\n= (no.of ways in which the first digit can be chosen) * (no.of ways in which the second digit can be chosen) * (no.of ways in which the third digit can be chosen)\n= 9 * 10 * 8\n==> m(a) = 720\n\n[Since the first two digits can be any of the ten and the last should be different from the first two, there would be 9 possibilities for choosing the first digit (since 0 is excluded), 10 possibilities for the second (any digit can be chosen including 0) and 8 for the third (excluding the ones chosen for the first and the second)\n\nProbability that the 3rd digit is different from the first two\n==> Probabilty of occurance of event "B" = m(b)/n\n==> P(A) = 720/900\n==> P(A) = 4/5\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/
457875
Boy, this is tricky.  I think the probability is (8! * 41c2 * 10c2)/49!, where 41c2 = 41!/39!2! and 10c2 = 10!/8!2!  In this number, 8! represents the correct picks, 41c2 the incorrect picks, and 10c2 the number of ways to order the picks.\n\nFor exactly seven correct numbers in 10 picks , only a minor change is needed.  It becomes (7! * 42c3 * 10c3)/49!.  Follow this pattern for 6 of 10, 5 of 10, etc
375916
diidy, how could the odds get worse when you buy more tickets?\n\nI think there are actually two answers, depending on whether each ticket is independent of the others or not.\n\nIf each ticket is independent, and there's no guarantee as to how many tickets there are or how many winners there will be, then think of it this way.  The odds of any ticket NOT winning are 299:300.  Therefore the odds of NO ticket winning are 299^n:300^n, where n is the number of tickets.  The odds of at least one ticket winning are 1 - the odds of no ticket winning, or 1 - 299^n:300^n, or (300^n - 299^n):300^n.\n\nIf there are 300 tickets of which 1 is a winner, then the odds of winning are n:300.
618540
It is called the Day-Month-Century Birthday Coincidence.\n\nEveryone born on January 9th in the 20th century (1900s) will experience the same occurrence. 1/9/00  to 1/9/99.\nThe next time this happens will be in the 22nd century (2100s). Those born on February 1st in the year 2100 to 2199 will experience the same occurrence.\n\nIt doesnt stop there. \nFrom 2200 to 2299, February 2nd is the special day of the year\nFrom 2300 to 2399, February 3nd is the special day of the year\nFrom 2400 to 2499, February 4nd is the special day of the year\nFrom 2500 to 2599, February 5nd is the special day of the year\nFrom 2600 to 2699, February 6nd is the special day of the year\n...\nFrom 9900 to 9999, September 9th is the special day of the year\n\nYears that are skipped: 3000, 4000, 5000, 6000, 7000, 8000, 9000
146232
The chance of you having the correct door is one third.  That's because when you first chose a door you had a one third chance of chosing the correct door.  So you had a two thirds chance of being wrong.  After you choose a door, at least one of the remaining doors will be an incorrect door.  This is the door that springs open.  It doesn't change the fact that you have a one third chance of having the correct door.  \n\nIf the door you picked is incorrect, the remaining door is correct.  The odds that you've picked a wrong door are two thirds, therefore, if you switch doors, the odds are two thirds that it will be the correct door.\n\nIt's not intuitive, but I think it's correct.\n\n\nI figured out how to make it intuitive!  Suppose you have 1000 doors.  You pick one.  You'll almost never pick the correct door.  Then, have 998 of the remaining doors opened, and shown to be incorrect.  If you switch, you'll almost always (99.9%) pick the correct door.  \n\nWith 3 doors, the same principal is involved, it's just not as extreme or intuitive.
535461
I understand what you are saying when you mention the permutations for 10 digits, but I think you're missing what it really means.  First, this isn't right to think of it this way to think about this.  Non-repeating doesn't mean that I can't find a string in the decimal expansion that's the same as an earlier string (for example, finding a piece 14159 later in pi isn't impossible).  What it means is that there is no point where the digits in pi *completely* repeat (i.e. it becomes something like 141591415914159.... on and on forever).  So you're right, it can include some kind of partial repeat, but non-repeating means it can't have a total repeat.\n\nYour comment about the permutations (not combinations) on the numbers {0,...,9} is right but it's used incorrectly also.  It's wrong because of what I mentioned above.  Basically it's assuming that once you pick a number you can't repeat it again in a 10 digit sequence, but that's clearly wrong (again, 14159 appears in pi, where 1 occurs twice).  So you have several right ideas here but you're combining them seomwhat incorrectly.\n\nHope this clears things up!!
661773
Let's deal with "b" first because it makes the general solution clear.\n\nFor an unbiased coin the probability of "Heads" (H) = probability of "Tails"(T) = 1/2 = 0.5\n\nIf n = 3, then you interested in the probabilty of two people having heads and one tails; or two people having tails and one heads.\n\nThe pobability of a single event where two people have heads and one has tails = 0.5 x 0.5 x 0.5= 0.125. But this can happen 3 ways: THH, HTH, HHT and therefore the probability that 2 people have heads and one has tails = 3 x 0.125 = 0.375. Of course the probability that two will have tails and one heads will also be 0.375. So the probability that a leader would have been chosen after one iteration (vote) = 2 x 0.375 = 0.75. (Note: if 3 coins are tossed there are only 4 possibilities: all heads, all tails, two tails and one head, or two heads and one tail.)\nThe probability that a leader is not chosen the first time = 1-0.75 = 0.25. And the probablty that a leader will not be chosen after two iterations is (0.25)^2 = 0.0625. And therefore the probabilty a leader will be chosen within two iterations = 1-0.0625= 0.9375 or nearly 94%. Now we'll deal with the general case.\n\na) For arbitrary n, and a single iteration, the probability of choosing a leader = 2n(0.5)^n. And the probability of not choosing a leader is 1-2n(0.5)^n. There is a difference between the probability of choosing a leader "in" the kth iteration and "by" the kth iteration. The probability that a leader will not have been chosen after k iterations is \n(1-2n(0.5)^n)^k. And the probabilty a leader will have been chosen is simply 1 minus this value.\n\nSample calculation: assume n = 20 and k=50\n\nProb. leader not chosen = (1-40(0.5)^20)^50 = 0.9981\n\nAnd the probability a leader will have been chosen = 1-0.9981 = 0.0019 which is less than 0.2 percent or 1 in 500. This shows that as n increases, the probability that one person will have a different result than everyone else becomes very small and enormous number of iterations would be required to have a reasonable chance of choosing a leader.
263333
1)  I think you're missing a bit.\n\nGiven that P(W<=4)=0.825 then P(W>4) = 1-0.825 = 0.175\n\nThis is because all probabilities sum to 1.  So since <=4 annd >4 make up all teh possibilities the value could take - then they're total probabilties must be 1.\n\n2)  Expected value\nTo do these you simply have to multiply the probability by the value and total them up.\n\n+36,000×0.7 -6,000×0.3 = $23,400\n\n3)  Binomial\nYou have a bin(6, 1/6) as there are 6 trials and the probability of a success (ie getting a four) is 1/6.\n\nYou require the probability of 2 fours so just using the binomial formula P(X = x) = nCx × p^x × (1-p)^(n-x) you get:\n\nP(X = 2) = 6C2 × (1/6)² × (5/6)^4 = 0.2009\n\n4)  Binomial\n\nthis again is a binomial Bin(10, 0.33) - as we have 10 trials (students) and the probability of each one being a success (ie member of ethnic minority) is 0.33\n\nUse the formula above to get the probability of more than 6:\n\nP(X = 7) = 10C7 × 0.33^7 × 0.67^3 = 0.01538...\nP(X = 8) = \nP(X = 9) = \nP(X = 10) = \n\ntotalling these up gives the probability of more than 6.\nThen the probability of NOT more than 6, P(X <=6) = 1 - P(X>6) using the idea from 1.\n\n5)  The mean of teh binomial is just np (ie the number of trails time the probabilty of success).\n\nSo for your example where we have bin(6, 0.7) it will be 6×0.7 = 4.2 seeds germinating.
667456
first of all, you made a typo:  it should be "in our house." and not "in your house."  The reason is, it is definitely not true for my house, so I have to assume that it is true for yours :).\n\nNow, what you did is you told us that the sum of the divisors of 36 (cause that is what the ages are) add up to some number.  You also gave us more information, because even if we did know the number of windows in your house (the sum of the divisors), there are two different answers possible.  Since 13=2+2+9=1+6+6 is the only number with two possible solutions, it must be this one.  Since you have an eldest child, we can assume that you don't have two eldest children (and thus don't have twins of age 6).  Thus you have an eldest child of age9  and twins of age 2.
58530
Mine are 1 and 3 which together are 13.  I was born on the 13th and everyone seems to think it's an unlucky number but I love it because everyone seems to stay away from it.  It always pops up, such as right now when I scrolled down for questions and yours was asked 13 hours ago!
539023
I would start by asking the kids questions.\n\nHow many days are in a year?\n\nSo, if we have 80,000 people how could we estimate how many people will be at each party? At least 1 kid will be smart enough to tell you to divide it. Be sure that write it on the board as you go.\n\nWhen you come up with 219 then ask the kids if they really think that the birthdays will be divided equally. If that is true, then if there are 24 kids in your class... there should be 2 birthdays in each month. (Maybe as a starter for the lesson you could have some sort of birthday poster and could quickly write the names of the kids under each month that their birthday falls. You may have some months with 5 birthdays, and others with 1 or none). If they say that they think that the birthdays will be evenly then use your birthday chart and explain that their birthdays are random; that it does not work out to exactly 2 birthdays in each month.\n\nThen you can explain to them that with 80,000 people there are likely to be a few birthdays on more days than others. They will get it; especially when you use their own data to help explain it.\n\nIf time allows you could do another problem that is similar, but maybe using something like the day they got their first pet (I don't know; something...) anyway, because some people may have never had a pet you could ask if it is possible that some people never come to celebrate. Or maybe use vet patients for the problem because then we could assume that all patients would qualify.
465914
I am! And so is my cousin......born on the same day but different years. She accuses me of stealing her birthday. I don't think the date is unusual but I believe that more people have summer birthdays because their parents planned it. Really, people say that. Kinda weird....
417130
luv phy is doing the calculations as if each letter could be placed in any one of the 8 envelopes, even after some of the envelopes have been used for other letters. In other words, he is treating the placing of the letters as 8 independent events. But they are not independent, as each event is affected by what has happened before.\n\nThe probability that no one will get the write letter is equal to the number of ways that can happen, divided by the total number of ways that the 8 letters could possibly be arranged. The second number (the denominator) is easy: it's 8! = 8x7x6x5x4x3x2x1 = 40,320.\n\nTo get the numerator, you have to do a calculation. There may be a standard formula, but one rule that works is this:\nThe number of ways that n envelopes can all contain the wrong letter (call this number W(n) ... "W" for "wrong") is\n(n-1) x (W(n-1) + W(n-2)).\nThis applies when n is 3 or more.\nFor n = 1, W(1) = 0 (i.e., you can't screw up if there's just one letter and one envelope).\nFor n = 2, W(2) = 1 (i.e., there's only one way to screw up 2 letters: you get them reversed).\n\nSo, applying the formula, \nW(3) = (3-1) x (1 + 0) = 2\nW(4) = (4-1) x (2 + 1) = 9\nW(5) = (5-1) x (9 + 2) = 44\nW(6) = (6-1) x (44+ 9) = 265\n\nI'll let you work out the values for W(7) and W(8).\nThen you need to divide W(8) by 40,320 to get the probability requested in (a) of the problem.\n\nOnce you solve (a), then (b) is simple. It's just 1 - (a). Either no one gets the right letter, or someone (at least one person) gets the right letter. So the answers to (a) and (b) have to add up to 1.\n\nGood luck!
620360
amazing coincidences.. maybe.. ur lucky number is 3!! who knows.. try to give it a shots~
92715
IMO, fb2kool is on to something.  Luck is used to describe a short series of events that appear to deviate from the average (when they simply contribute to the average).  The strange thing about probability is that the "average" never seems to happen.
150114
I was one of the lucky ones.  There were 4 of us born at Fort Devens Army Hospital in Shirley Massachusetts.  Growing up I would celebrate 1 day with family and 1 day with friends.  If there was a 29th then I'd get 1 day only.  Now it seems like it doesn't matter if there is a 29th or not, my birthdays seem to last for a week.  I have had my ID torn up because the bouncer thought it was a fake.  He said there was no Feb 29.  The bar paid for a new ID.  The fun part is when I can say my kids are older than me.  By the way...I turned 10 1/2 this year.  I'll be 11 in 2008.  I sometimes get gag gifts from people.  Like when I turned 6 (24) I got a barbie doll and a card for a 6 year old.  All in all, it's just another day though.