wut is the probability that a family with 3 children wil have exactly one girl Define a trial 4 the simulation
398876
it's 50 50 every time.  genetics states that you get one chromosome from your mom (always X) and one from the dad, X or Y.  XX means girl, XY means boy. (assuming dad has even amounts of both).\n\nHence you use condition probablity assuming each occurance is a 50% chance.\n\n50% x 50% x 50% = 12.5%
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.
171633
probability is a number from 0 to 1 (or 0% to 100%) that describes how likely an event is to occur\nexample~ jenny had 10 marbles. 3 were red. 6 were blue. and 1 was white. she put them all in a bag. she randomely drew one out of the bag. what is the probability that she will pick blue?\n            answer= 6/10 which deducts to= 3/5\n\nactual events is what really did happen. \n\nprobability is guessing based on passed expirements or facts. actual event is a fact.
388148
All seniors with at least 2 girls means either 2 girls and 2 boy or 3 girls and 1 boy or 4 girls and 0 boys.  There are not 4 senior girls so the last cannot occur.\nC(3,2)C(3,2) = 3*3 = 9\nC(3,3)C(3,1) = 1*3 = 3\n9 + 3 = 12\nThe total possible ways of picking 4 people is\nC(12,4) = 495\n12/495 = .024242\nb)\nmore girls than boys means 3 girls and 1 boy or 4 girls and 0 boys\nC(7,3)(C(5,1) = 35*5 = 175\nC(7,4) = 35\n175+35 = 210\n210/435 = .424242
600596
There are 6 possible outcomes for the roll. There is only one outcome when the number is 4. Therfore the chances of getting a 4 are 1/6.
548638
Assuming you'd've had two kids anyway, 1 in 365. Much less if you only ever have sex on September 20th ;c)\n\n[edit: 1 in 365 is the odds of the two sharing *any* particular date. The odds of sharing that one particular date is 1 in 365 x 365 = 133225]
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
435037
try this site for help on probabiliity\n\nhttp://www.futureaccountant.com/probability/
874576
Note: they aren't asking just for the chance that at least 2 children have blue eyes.  They are asking for the probability, given that you know at least one already does.\n\nThis is conditional probability.\n\nP(B | A) = P(A and B) / P(A)\n\nSo the probability that you have at least 2, given you have at least 1 is the probability you have at least 2 divided by the probability you have at least 1.\n\nFor that part though, you would use binomal distribution.\n\nFirst, what is the probability that at least one child has blue eyes?\nThis would be 1 minus the probability that nobody has blue eyes.  \nP ( blue >= 1 ) = 1 - (0.75)^5 \nP ( blue >= 1 ) = 0.762695312 \n\nNow the probability that at least *two* children have blue eyes:\nP( blue = 2 ) + P ( blue = 3 ) + P ( blue = 4 ) + P ( blue = 5 ).  Again you can figure the negative statement and subtract from 1.  So this would be equivalent to:\n1 - P ( blue = 0 ) - P ( blue = 1 ).\n\nThis would be:\nP ( blue >= 2 ) = 1 - (0.75)^5 - (0.25)(0.75)^4 * C(5,1)\nThe C(5,1) means the ways to choose 1 person from 5, which is just 5.\nP ( blue >= 2) = 1 - (0.75)^5 - (0.25)(0.75)^4 * 5 \nP ( blue >= 2) = 0.3671875\n\nSo putting it all together:\nP ( blue >= 2 | blue >= 1 ) = P ( blue >= 2 ) / P ( blue >= 1 )\n= 0.3671875 / 0.762695312\n= 0.481434059\n≈ 48.1%
220111
Greetings:\n\nHaving selected one of the four, the probability that person #2 will have a different date of birth is 364/365 (if we seek no repeats, #2 can have any 364 of 365 possible b.days).  Similarly, having selected two, the probability that #3 will have a unique d.o.b. is 363/365 (any 363 of 365 possible dates okay).  Finally, for #4, the probability is 362/365.  Hence by the multiplication property, the probability that all four will have different b.days is given by, (364/365)(363/365)(362/365) = (364)(363)(362)/(365)^3 = 0.984 to three places.\n\nRegards,\n\nNikkor180
812841
Given 5 children there are 2^5 or 32 possible combinations of girls and boys.  Since there is only one combination with all boys that leaves 31 combinations containing at least 1 girl.  Therefore the probability is 31/32.
731902
There are 20 students in the class. There are 4 possibilities when you choose the 3 random students. I've included the probabilities of these 4 possibilities. 1) and 3) are the answers to your questions.\n\n1)   3 boys = (4/20)(3/19)(2/18) = 0.0035\n\n2)    2 b, 1 g = (4/20)(3/19)(16/18)x3(=3C2) = 0.0842\n\n3)    2 g, 1 b = (16/20)(15/19)(4/18)x3(=3C2) = 0.4211\n\n4)     3 g = (16/20)(15/19)(14/18) =  0.4912\n\nFor the probabilities to be correct they must sum to 1:\n\n0.0035+0.0842+0.4211+0.4912 = 1
1028806
I'll assume you already know what permutations and combinations are and how they work.  Now lets say you're conducting an experiment where the result of the experiment can either be a success or a failure (this is called a Bernoulli trial in probability).  To ensure the accuracy of your results, you are going to repeat the exact same experiment three times, and you will look at the total number of successes in those three Bernoulli trials.  If we know the probability that each individual trial will be a success, we can calculate how many successes we can expect to get in three trials.  This is called a binomial distribution.  Let's look at all possible outcomes:\n\nYou could have all successes:\ns s s\nYou could have all failures:\nf f f\nThe first trail may be a success and the other two failures:\ns f f\nAnd so on... here's the list of all possible outcomes:\ns s s\ns f f\ns s f\ns f s\nf s s\nf s f\nf f s\nf f f\n\nLet's say you want to know what's the probability that you will get exactly two successes and one failure from your three trials.  The number of possible outcomes from this list that have exactly two successes in it is (3 choose 2)=3 out of the eight in the list.  Now if "p" is the probability that each individual trial is a success, and if "q" is the probability that it is a failure (meaning q = 1 - p), then the density function for this binomial distribution is:\nf(x) = (3 choose x) * p^x * q^(3-x)\nSince we want to know the probability of getting exactly two successes, we plug in two for x:\nf(2) = (3 choose 2) * p^2 * q^(3-2)\nf(2) = 3 * p^2 * q\nDepending on what p and q are, the probability of getting two successes in your three experiments can change.\n\nThe binomial distribution is one example of many discrete distributions that use combinations in its formulas.  Permutations have other applications, but this answer is getting long, so I'll leave it at that.
90357
F- even and odds.  There are 3 even sides and 3 odd sides. So the likelihood is the same for both.
694491
each die has 6 ways it can land\nonly one of them is a 1\nso chance of one die getting 1 is 1/6\nsame with the other one\nSo the odds of rolling "snake eyes" is 1 in 36\n\nchance of being a girl is 17 out of 30 or 17/30\nchance of being an A student is 5 out of 30 (5/30) or 1/6
727752
Number of all permutations = 8!\n\nNumber of all permutations with 5 boys first (then 3 girls) = 5!3!\nNumber of all permutations with 5 boys last (3 girls first ) = 5!3!\n\nSo the probability in question is 2*5!3!/8! = 1/28
557990
This is a binomial problem.  The number of trials is 4 and the probability of success is 1/6.  No more than twice means either 0, 1, or 2 sixes.  So\n\n4!/(0!4!)(1/6)^0(5/6)^4 + 4!/(1!3!)(1/6)^1(5/6)^3 + 4!/(2!2!)(1/6)^2(5/6)^2\n\n= 0.482 + 0.386 + 0.116\n\n= 0.984