four out of nine puppies are black what is the probability that two are black?
345957
What are the number of puppies you have chosen ?\n\nTry this site for help\n\nhttp://www.futureaccountant.com/probability/
67102
43 out of 400.\nThere is 120*120 pairs of minuites within which in which they can arrive resturent. And to be within 3 minuite, the diffrence of this of this  two pair have to be 3 or less. If the first one of this is between 4 to 117, there will be 7 possibility of  second one.\nAnd teher will be 6 for 3 and 118, 5 for 2 and 119,and 4 for 1 and 120.\nso total possibolity is 114*8+6*2+5*2+4*2 that is 828\nSo the probability is 828/14400=43/200.
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?
97181
half \nedit read question wrong it 2/3 of black will be heter\n\nOkay the question is What fraction of the BLACK f2 is expected to be heter And since 1/4 is Homozygous Black and 1/2 are heterzygous black.. so 2/3 Of the BlACK f2 will be heterzygous.
968185
First the probability that a family of 4 is all girls is:\n1/16 (and the chance that they are not is 15/16).\n\nRather than figuring the chance of at least one family having this arrangement, figure the probability that *none* of the families do.  Then subtract from 1.\n\nThe answer will be 1 - (15/16)^10\nor about 47.6%\n\nIt's not a simple multiplication by n... if it were, then with 16 families you should have a 100% chance.  And with 32 families you would have a 200% chance.  That logic would be correct if you couldn't repeat a combination, but you can.  Just because family A has four boys, for example, doesn't mean family B can't also have four boys.\n\nLabel the 10 families as A, B, C, D, ... , I, J.  The probability that A has four girls is 1/16.  The probability that B has four girls is 1/16, etc.\n\nThe chance that *none* of them have four girls is 15/16 x 15/16 x 15/16 x ... x 15/16.  This is (15/16)^10 or about 52.4%.  Therefore the chance that at least one of them has four girls is 100 - 52.4 or 47.6%
906595
C.\nThis is a binomial probability problem.\nThe binomial formula says \nP(r out of n) = (nCr)(p^r)(q^(n-r)), where n is the number of trials (here 6 rolls), r is the number of successes (here 0, 1, 2, and 3 twos), p is the probability of success (here 1/6), and q is the probability of failure (here 5/6)\nYou need to plug in 0, 1, 2, and 3 for "r" into this formula.  If you do, you get:\nP(0 twos) = .3348979767\nP(1 two) = .401877572\nP(2 twos) = .200938786\nP(4 twos) = .05358367627\nYou add up these answers to get your final answer, about .9913
954146
B- 8 pups.  The other 3 examples are of ordinal numbers (first, second, third, etc.)  Here's a testing tip you might like to pass on to your niece.  If 3 out of the four answers are similar, as they are in this case, your best bet is to pick the fourth answer that is different from the rest.
1070800
none, only small brown dogs are born
435037
try this site for help on probabiliity\n\nhttp://www.futureaccountant.com/probability/
389934
A lab and a collie could get the award in 7 times 8 or 56 ways.\nA lab and a sheltie could get it in 7 times 2 or 14 ways.\nA sheltie and a collie could get it in 8 times 2 or 16 ways.\n\nSo that's 56+14+16 or 86 total ways.\n\nOops, I didn't see that the awards were different.  Now the answer must be doubles to get 172 ways, since if Collie A got award 1 and sheltie S got award 2, that's different than if Collie A got award 2 and Sheltie S got award 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
425689
8 dogs with 8 puppies that is 64 puppies.\n\nSo 8 dogs have 4 legs each - 32 legs 32 paws\n64 puppies with 4 legs each 256 legs 256 paws\n8 humans with 2 legs each 16 legs 0 paws\n\n304 legs in total\n288 paws\n\ndon't understand the 8 men and each 8 has sacks? if you mean each man has 8 sacks with 8 dogs then\n64 dogs with 4 legs each 256 legs / paws\n512 puppies with 4 legs 2048 legs / paws\n\nso total 2420 legs, 2404 paws
925058
Assuming Black is dominant (B) and Albino is recessive (b)\n\nThe first cross produced all black\nbb x BB\n---B-----B\nb Bb----Bb\nb Bb----Bb\nFrom here, you can see that all the offsprint have one dominant Black gene, so they will all be black.\n\n\nThe second cross had about half albino\nbb x Bb\n---B-------b\nb  Bb-----bb\nb  Bb-----bb\nFrom this one, you can see that half of the offspring are homozygous recessive bb, and are albino.\n\n(Ignore the ---- in hte cross, they don't mean anything, they are just for spacing because yahoo deletes multiple spaces)
1115185
25% chance of being O\n75% chance of being A\nonly if both parents are heterozygous\n\neven if one is homozygous no chance of getting "O"
568866
is there the same probability it will go into each bin?\n\nthen 350/9=38.8888 so about 39 in each
10223
Puppies are always the answer.