how to find a formula of the probability that in 20 people, at least two of them have the same birthday?
548980
Here's the formula:\nThe probability that all birthdays are different is\nP(365,r)/365^r, where r is the number of people and P stands for permutation.\nThus, the probability that at least two are the same is\n1 - P(365,r)/365^r\nr = 20\n1 - P(365,20)/365^20\n= 1 - .588561616419\n= .411438383581\nSo there is a .4114 chance that at least 2 people in a group of 20 people will have the same birthday.\nThe birthday problem is pretty interesting.  If there are 23 people, the probability that at least two have the same birthday is .507.  If there are 50 people, it is .970.
937552
You can do this two ways...\n\n1. Find P(x=1) + P(x=2) + P(x=3).\n\n2. An easier way: find P(x >= 1) by taking P(x = 0) and subtracting that from 1.\n\nEither way, from the table, you should get the same thing.\n\nedit: That rule of thumb will work for the binomial distribution because it is discrete.  It would not for, say, continuous distributions.  A better rule of thumb might be this...\n\nP(X >= x) = 1 - P(X < x).\n\nThis will always work.  Hopefully you understand the reason behind it; the probability of an event is equal to one minus the probability of the opposite of that event.
366018
try this site\n\n\nhttp://www.futureaccountant.com/probability/
631161
This is a binomial probability.\nP(success) = 1/5, P(failure) = 4/5.  There are 6 events here, so P(2 successs) = (1/5)(1/5)(4/5)(4/5)(4/5)(4/5).\nP(at least 2 successes) = P(2 successes) + P93 successes) + . . . . P(6 successes) or 1 - P(O successes) + P(1 success) + P(2 successes) which you already computed above.\nLet me know if you don't understand.
258297
Each game you have a 50/50 chance of being right, so that's 1/2^1(^is the sign for an exponent), there are 32 first round games, so for picking the perfect first round the chance is 1/2^32, for the second there are 16 games, you multiply the odds of picking these right (1/2^16) and continute for the number of games each round, so your formula will be, 1/(2^32*2^16*2^8*2^4*2^2*2^1) which will equal 1/9,223,372,036,854,775,808.  There is 9,223,372,036,854,775,808 different combinations of winners, and only one is correct for the perfect bracket.
192695
You have the following lists:\nB B B\nB B G\nB G B\nG B B\nB G G\nG B G\nG G B\nG G G\n\nSo you have 8 variations and of those 3 are 2 boys 1 girl so the probability is 3/8.
430637
First choose which number it is.  There are 6 numbers, so there are C(6,1) = 6 choices.\nThe probability of that number is 1/6\nSince there are three\n(1/6)(1/6)(1/6) = 1/216\nNow multiply this times 6\n6/216 = 1/36
815499
The correct answer is 15, but the proper formula is (6C4)/2\n\nBy the way, you can check this by figuring out how many groups of 2 you can have, since every group of four being chosen is the same as a group of two not being chosen.  For the first person, there are six people to choose from.  For the second, there are five.  However, since picking Person I and Person II is the same as picking Person II and Person I, you have only half as many possible combinations, or 15.
138694
each child can be a male or a female \ntherefore the total no of events possible is 2^6=64\n\nthe question says that you should have atleast 3 girls \ni.e there can be three, four ,five or all six girls .\n\nthe number of ways of choosing 3,4,5,6 girls from 6 children is the sum\n6C3 + 6C4 + 6C5 + 6C6 (here C stands for combinations )\n\ni,e\n\n(6*5*4)/(3*2*1) + (6*5*4*3)/(4*3*2*1) + (6*5*4*3*2)/(5*4*3*2*1) + 1\nthis reduces to \n20+15+6+1\n=42\n\ntherefore the probability is \n42/64
840938
The number of handshakes is 302C2, i.e. (302)(301)/2 = 45451.\n\nYou're essentially looking for the total number of pairs of people in the group.  With your answer of 90902, I suspect that you were on the right track.\n\nSo to find the number of pairs, you have 302 choices for the first and then 301 choices for the second, making 90902 possibilities.  We have to cut that in half because this is counting 'pick person A then pick person B' and 'pick person B then pick person A' as two distinct choices, but this is really just two ways of looking at the same handshake.  Cheers!
856461
This would be a binomial.  Assuming that p=1/2 , equal probability of male or female.\n\nThen it's like 80 trials with success (male) or failure (female)\n\nAt least 30 male means that you have to add probs of 30, 31, 32, ...,80.  It's easier to add probs of 1,2,...,29 and subtract from 1.\n\n80 is a large enough number to use the normal approximation.\n\nFor the normal, the mean would be 80/2 = 40.  So, at least 30 is\n\n1-P(X<less than 30) \n\nconvert X to Z to use standard tables.
584200
1) the pdf is actually a binomial.\nIf you consider only 1 man, what is the probability that he will be paired to his wife out of the 10 women? \nIt is 1/10\n\nLet X be the number of paired couples, paired meaning they are paired to their own spouse.\nTherefore X~B(10,1/10)\n\nSo if we want to find the probability that there is only 1 pair, using the formula of a binomial dist,\n\nP(X=1)= [10 choose 1] (1/10) (9/10)^9\n= 0.3874\n--------------------\n\n2) To find the mean and variance, using the Binomial parameters formula, we compare what we got in part 1 with the general formula,\nX~B(n,p)\nX~B(10,1/10)\n\nSo our mean = (n) multiply (p)\n= 10 x (1/10)\n= 1\n\nVariance = (n) multiply (p) multiply (1-p)\n= 10 x (1/10) x (9/10)\n9/10\n----------------------\n3) P(X >= 3) = 1 - P(X <= 3) \n= 1 - [ P(X=0) + P(X=1) + P(X=2) + P(X=3) ]\nreferring to the above eg in part 1, its easy to derive\n= 1 - [ 0.0349 + 0.3874 + 0.1937 + 0.0574]\n= 0.3266
245215
There are five tosses with two choices each.\nEach toss is a 50% chance of coming up heads regardless of the results before it or after.\nThe probability of getting a certain number of heads would be calculated as such: number of heads/number of choices.\nWe are looking for two heads and there are ten choices (heads and tails of each coin)\nThe probability of it would be 2/10 or 20%\n\nIt is correct in that each toss is 50% chance, but in this case we are speaking of of the chances of a group of coins having a certain number turn up heads.  We are also speaking of the probability of this before any actions are taken.  The odds will change with each coin toss till the last one which will be 50%
365970
1/161\nTake the number of ways the event can occur divided by the total number of events.
324794
its easy. figure out in this way there are 3 digits and since its an odd number the one's place or the last digit can be 1,3,7,9(5 is excluded).the no. of chances are 4. for the second digit the digits can be 0 to 9 leaving 5 so no. of chances are 9. for the first digit 0 and 5 are not possible and the no. of chances are 8.\n\nmultiply the no. of chances for every digit it ll give u the answer i.e 8*9*4 = 288. so u have 288 different numbers.
492063
Count the total number of groups:\n9 choose 5 = 9*8*7*6/(4*3*2*1) = 9*2*7 = 126.\n\nb) Count the number of groups where 3 women where chosen and 2 men.\n(4 choose 3) * (5 choose 2) = 4 * 10 = 40.\n\nThe prob that a random committee has exactly 3 women = 40/126\n\nc) If there were no women then the number of ways this can happen is:\n(4 choose 0) * (5 choose 5).  As you could probably already see, there is only 1 way to have a committe of 5 men and 0 women.  The prob = 1/126.\n\nf) >=3 women is the same as saying 3 or 4 women as 4 women is the most you can have.\n\n(4 choose 3)*(5 choose 2) + (4 choose 4) * (5 choose 1)=\n4*10 + 1*5 = 45.  Prob = 45/126.
91512
20 questiosn with five options, the probabilty to choose the correct one is 1/5 per each question.\nThe probability to answer all 20 questions is (1/5)^20.\nThe probability to answer more than 10 is (1/5)^11\nThe probability to answer any eight is (1/5)^8\nThe probability to answer maximun 7 is (4/5)^13