a couple plans to have 5 children.  find the probability of having at least one girl?
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.
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
860554
27 ways to pick the first person in the group\n26 ways to pick the second\n25 ways to pick the third\n24 ways to pick the fourth\n23 ways to pick the fifth.\n\nHowever, order is irrelevant, so we need to divide by 5!.\n\nThis is 27 choose 5\n= 27 x 26 x 25 x 24 x 23 / 120 \n= 80,730 ways
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
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.
907371
I know of a family that had 4 kids......they were first cousins, and they ended up having 4 kids with learning disabilities....and downs syndrome......\n\nSo I would say no...do not have kids unless you are prepared to have kids with many challenges.....
1162527
Because you are looking for the number of girls in relationship to the total class, you have to figure out what proportion of the total are girls. Out of every 9 students, 4 are boys and 5 are girls, so 5/9 of the students are girls.\n\nG/45=5/9\n\nSolve for G.
849665
The first two answers are the same answer ... 1/16 = 0.0625 ... and both are incorrect.  (no_life computed the odds that the first child only is the male ... but forgot about the cases where the second child is the only male, or the third child, or the fourth.  And Steph computed the odds that *all four* are boys ... in which case, yes it is 0.5*0.5*0.5*0.5.)\n\nThe correct answer is 1/4  = 0.25 = 25%.\n\nHow many possible combinations can you have?\n2x2x2x2 = 16\n\nHow many of them are "one boy + 3 girls"?\n4 ... the first child is a boy (and the rest girls), or the second child, or the third, or the fourth.\n\nSo the odds are 4/16 = 1/4 or "1 in 4" (or "4 to 1 against").
806831
All your responders so far have assumed that the next child's gender is independent of the prior children's genders. That is a correct assumption.\n\nThey have also assumed that the probability of having a boy is 50%. This is not quite true. Population-wide, the ratio of baby boys to baby girls is approximately 105 to 100, so the percentage is actually a bit over 51%. \n\nThis is also based on the assumption that the probabilities do not vary from one couple to another. You may have read descriptions of techniques for increasing the probability of having a girl, or of having a boy. And if so, you know that the things you have to do (I won't get too graphic here) are things that some couples are likely to be doing already (because of the geometry of their bodies, or the chemistry of their bodies, or because of their technique). So I suspect (and this is only my opinion) that the probability of having a child of one gender or the other varies from couple to couple.\n\nThat last-mentioned suspicion could probably be tested by determining the relative frequency of families with different combinations of boys and girls. Do siblings of the same sex tend to occur more frequently than if each birth were a 51/49 proposition? If so, then there is likely a variation from couple to couple. And if they occur LESS frequently, then we REALLY have a conundrum, and I have NO idea what it would mean.
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
651368
Answer: 4\n\nPut 30 kids in each room. So, after 10 rooms are filled, you have out (30*10) kids in the 10 rooms. So, you are left with 6. If you put 1 kid in 6 rooms, to make the number of kids in these 6 rooms to 31, you are left with 4 rooms with 30 kids.
783821
5 out of the 8 items in Box A are non-defective.\n3 out of the 5 items in Box B are non-defective.\n\nThe probability that an item from Box A and an item from Box B are drawn and are both non-defective is:\n\n5/8 * 3/5 = 3/8\n\nThe probability that an item from Box A and an item drawn from Box B and both defective is:\n\n3/8 * 2/5 = 3/20.\n\nThis means that the probability that one of the two items drawn is defective is:\n\n1 - p(both are not defective) - p(both are defective)\n=1 - 3/8 - 3/20\n= 19/40\n\nWe can see this by showing all possible outcomes. The items in Box A are shown on the top edge, and the items in Box B are shown on the right edge\n\nN  N  N  N  N  D  D D\nA  A  A   A  A  B  B  B  N\nA  A  A   A  A  B  B  B  N\nA  A  A   A  A  B  B  B  N\nB  B  B  B  B  C  C  C D\nB  B  B  B  B  C  C  C D\n\nThe events marked "A" represent 2 non-defective items. The events marked "B" represent 1 defective and 1 non-defective item. The events marked "C" represent 2 defective items.\n\nOut of 19 events where there is 1 defective and 1 non-defective item, the defective item came from box A 9 times.\n\nSo if 1 item is defective and 1 is not, the probability is 9/19 that it came from box A
861545
It depends on what kind of A the Mom is. If the Mom is AA, then 100% of the offspring will be AO. If the Mom is AO, then each offspring has a 50:50 chance of have AO or O.
594326
50/50 chance it will be a boy for each birth therefore three boys
415287
Although this seems to be a problem about Gregor Mendel and his tall and dwarf pea plants, once it is realised that the probability of any child being short is 1/2, it simplifies to being the simpler and more familiar one of tossing a coin, heads or tails, n times\n\nIt would have produced more interesting results perhaps if the parents were Aa and Aa and three children out of four were therefore tall and we had looked at grandchildren (offspring of children who are one quarter AA, one half Aa and one quarter aa, eg if they intermarry with another such distribution among their partners.) However ...\n\n(i) is imprecisely worded. How big is the family?\n\nIf there is one child, the chance of that child being short is  1/2\n\nIf there are two children, the chance that both are tall is 1/4 and the chance that at least one child is short is 3/4. (Of one being short is 1/2 and of both being short is a quarter.)\n\nIf there are three children, the chance that all three are tall is 1/8 and the chance that at least one child is short is 7/8 (of one being short: 3/8, of two being short also 3/8, of all three being short: 1/8).\n\nWith 4 children, the chance of at least 1 being short is 15/16, of 1  being short is 4/16 or 1/4; of 2 being short is 6/16 or 3/8; of 3   being short is 4/16 or 1/4; of all 4 being short is 1/16;\n\nOne just reads across successive lines of Pascal's Triangle.\n\n(ii) the answer to this is buried in the above: 3/8ths\n\n(iii) again, imprecisely phrased. Does "such families" refer back to (i) or to (ii)? Taking it to mean families of four children, there is one chance in 16 of 4 tall children, 4 chances in 16 of 3 tall children, 6 chances in 16 of 2 tall children, 4 chances in 16 of 1 tall child and 1 chance in 16 of 0 tall children.\n\nIn 16 such families, there are 64 children and unsurprisingly 32 are tall\n(1 x 4) + (4 x 3) + (6 x 2) + (4 x 1) + 0\n\nwith an average number of 2 tall children per family, and as you would expect an average number of 2 short children per family, too.\n\n(iv) 256 = 16 squared and so the expectation is 16 families out of 256 will have 4 tall children, given 1 family in 16 has this particular expectation.
377837
well you have 2 options here.you could try asking her out an getting her in the sack and secretly try to go get the other in the sack and run back and forth between the 2 girls and pobably end up with 2 kids by different girls, or you could do the right thing,forget about girl#1 and take out girl# 2 and treat her with respect,love her and end up having a happily family someday\n\n\n\nit is your choice man--------go do whats right
564027
At least two of them, or exactly 2 of them?\nOf couse, assuming they choose just one recipe for "supper"\n\nProb. that at least 2 of them choose the same recipe.\n\n1 chooses a recipe:\nprob 2 picks diff. recipe = 7/8\nprob 3 choose diff than 1 and 2 = 6/8\nprob 4 diff than all above = 5/8\nprob 5 diff than all above = 4/8\n\nprob they all choose diff = 7/8 x 6/8 x 5/8 x 4/8 = 0.205\nprob at least 2 choose same = 1 - .205 = 0.795 or about 80%\n\nprob exactly 2.\n\n1 chooses recipe\n2 chooses same = 1/8\n3 chooses diff = 7/8\n4 chooses another diff = 6/8\n5 chooses another diff = 5/8\n\nprob 1 and 2 choose same (rest diff) = 0.0513\nhowever 2 and 3 (or any pair) could choose same (rest diif)\nnumber of possible pairs from 5 is (5)(4)/2 = 10\nSo probability that exactly two are same = 10 x .0513 = 0.513 or about 51%. \n\nnote: prob exactly 2 (51%) is less than at least 2 (80%)