Whats the probability of having at least 3 girls in a family of six children?
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
904590
from 5 women 3 women can be chosen in 5C3 = (5*4*3)/(1*2*3) = 10 ways\n2 men can be chosen from 4 men in 4C2 or 6 ways\n\nso total number of ways = 10* 6 = 60
549442
no, sorry. but there are a lot of people, men or women, who just think 5 kids are too many. but i'm sure theres someone who loves kids just as you do, and wont mind it. \nit probably also depends on the age of your kids.
594326
50/50 chance it will be a boy for each birth therefore three boys
395430
Assuming that the chance of having a girl is 50%, the chance would be 3/8.  This answer can be most simply reached a couple different ways.  First, you could list the possibilities:\n\nBBB\nBBG\nBGB\nGBB\nGGB\nGBG\nBGG\nGGG\n\n3 of the 8 possibilities fit the condition (exactly one girl), so the probability is 3/8.\n\nAnother way, better for larger numbers where the above method is impractical, is to use permutations.  The above works fine for this problem, though.
1009251
Ahh!  True, 6 consecutive flips of a coin in which each lands heads has a (1/2)^6 probability of happening.  But what are the odds that the seventh flip will be heads, too?
1141540
None I want to wait a while before I have any.Congradulations on your child any way,how old they!?But I want either 2 identical twin boys or just 1 boy or 1 girl thats it!
207532
7\nmy oldest daughter was born on 7-7-97\nand I was told I would I never have children!\n\nsecond favorite number would be 3, I now have 3 girls!  =0)
590754
6/16=37.5%\n\nSixteen different ways to have 4 children (excluding hermaphrodites.)\n\n1. b b b g  \n2. b b g b\n3. b g b b\n4. g b b b\n5. g g g b\n6. g g b g\n7. g b g g\n8. b g g g\n9. b b b b\n10. g g g g\n11. b b g g *\n12. b g b g *\n13. b g g b * \n14. g g b b *\n15. g b g b * \n16. g b b g *
704913
Suppose there are 13 girls and 11 boys in your class.  Then the ratio of girls to boys is 13 to 11 or 13:11 or 13/11
881626
Construct a punnet square:\n\nMom=  I^A & i\nDad= I^B & i\nOffspring will have a 1/4 chance of being: I^AI^B, I^Bi, I^Ai, ii\n\nIf the mother is I^Ai and the father is I^Bi, the possible blood phenotypes in the offspring are AB, B, A, and O.  There is a 25% chance of each blood type.
613390
i have one girl, would like 1 more girl and then one boy.
1098304
5 brothers and 2 sisters......4 older brothers and 2 older sisters...i am the BABY  GIRL and i have one brother under me....believe it or not, we all have the SAME MOTHER AND FATHER.... that's odd in today's world.......lol
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.
755941
4/16 or 25%\n\nThe follow table are the possibilities\n\nGirl    0   1   2   3   4\nBoy  4   3   2   1   0\n\nThe probability of having all boys is easy to figure out; it is 1/16. That is calculated by multiplying 1/2x1/2x1/2x1/2. The probability of having all girls is the same. 1/16.  \n\nThe probability of 2 boys and 2 girls is calculated by noting that 2 circumstances exist. In one the first 2 children are the same sex; the last 2 must be the opposite sex. Therefore you multiply  1/2x1/2x1/2 and get 1/8 (or 2/16) of this occurring. In the chance that  first 2 are equally split one boy one girl then the third child can be either sex and only the last one matters leading to another 1/2 x 1/2 = 4/16. For  a total of 6/16\n\nTherefore the remaining probably of 3 boys and 1 girl  or the alternative of 1 boy and 3 girls is the difference namely 8/16. \n\nSince it is equally probable to get 3B/1G or 1B/3G, the probability of 1 girl is half of the 8/16, namely 4/16
219821
This cannot be ascertained without knowing how many are being selected in all...\n\n\nhttp://www.futureaccountant.com/probability/\nhttp://www.schoolingkids.com/
531656
33% is equal to approximately 3 out of 10(3.3)\nTherefore, if six out of ten were to be selected, that would equal 66% of ethnics - more than twice the probability.\nIt follows then that in iself, the odds would be just over two-to-one against this happening.\nSo the probability of more than of no more than six would be two tone ON.\nDoes that make sense?