A bag contains 20 chocolates,15 toffees and 12 peppermints.If 3 sweets are chosen at random,?
456070
Chances of getting a non chocolate on:\n1 draw: 27/47 or 57.4%\n2 draw: 26/46 or 56.5%\n3 draw: 25/45 or 55.6%\n\n57.4% * 56.5% * 55.6% = 18%\n\nThanks Bogusfrog :)
931430
Fudge and whipped cream.\n1000
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%)
749946
a box of donuts..hmm donuts
656922
Okay, intially, you pick one marble, out of 12. and there are 3 yellows, then your initial chances are 1/4 of picking a yellow. Now, you've picked one, so there are 11 marbles left in the bag. There are 4 blue marbles in the bag, so you have a 4/11 chance of picking a blue ball at tthis point. \nTo Calculate Probability based  on an event that 1) Has already happened ( the yellow ball was picked first) and that is an event depending on the results of the , you simply multiply the chances. ( 1/4*4/11) or 4/44, or 1/11 chances that that will happen.
277836
multiply together the number of different possible combinations.  As the guy above me said, there are four times of toppings times the number of cheese choices times the weight of meat times the bun choices times anything else you can think of.
719411
you can enjoy so many styles of chocolate....i would say chocolate
247553
Total no of items\n = 7 forks + 4 spoons + 5 knives \n = 16  \n\nIn the experiment of drawing 3 items from among 16\n\nTotal no. of possible choices \n= no. of ways in which the 3 items can be drawn from the total 16\n==> n = C(16,3)\n==> n = 16 * 15 * 14/3 * 2 * 1\n==> n = 560 \n[On the assumption that the drawing is being made without replacement]\n\nLet \n"A", "B", "C" and "D" be the events of the three items drawn being 3 forks, 2 forks and a knife, 3 spoons and 1 fork, 1 knife and 1 spoon respectively\n\nFor Event "A"\n\n              Forks      Knives    Spoons     Total\nAvailable       7           5        4          16\nTo draw         3           0        0          3\nChoices       C(7,3)      C(5,0)    C(4,0)     C(16,3)\n\nNo. of favorable choices\n= No. of ways in which the 3 forks can be drawn from the total 7\n==> m(a) = C(7,3)\n==> m(a) = 7 * 6 * 5/3 * 2 * 1\n==> m(a) = 35\n\nProbability that the three items drawn are forks = m(a)/n\n==> P(A) = 35/560\n==> P(A) = 1/16\n\n\nFor Event "D"\n\n              Forks      Knives    Spoons     Total\nAvailable       7           5        4          16\nTo draw         1           1        1          3\nChoices       C(7,1)      C(5,1)    C(4,1)     C(16,3)\n\nNo. of favorable choices\n= (No. of ways in which 1 fork can be drawn from the total 7)* (No. of ways in which 1 knife can be drawn from the total 5) * (No. of ways in which 1 spoon can be drawn from the total 4)\n==> m(d) = C(7,1) * C(5,1) * C(4,1)\n==> m(d) = 7 * 5 * 4\n==> m(d) = 140\n\nProbabilty for the 3 items drawn to be 1 fork, 1 kinfe and 1 spoon = m(d)/n\n==> P(D) = 140/560\n==> P(D) = 1/4\n\nYou should be able to solve the two parts in between for Events "B" and "C".\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/probability/
830557
Tom baked 4 batches of fairy cakes. Each batch consisted of x cakes. He put 3 cherries on top of each cake in the first batch, but couldn't resist taking off and eating 5 of them. He did the same again with the second batch, then the third, then the fourth! Meanwhile, Sarah baked 5 batches of fairy cakes, again with x cakes in each batch. She put 2 cherries on each cake and an extra two on the final one. She repeated this procedure with each of the batches. She managed to resist eating any at all! Tom and Sarah discovered that Tom's cakes had exactly the same number of cherries in total as Sarah's. Set up an equation for this.\n\nThen: \n\nNumber of cherries on Tom's cakes = 4(3x-5)\nNumber of cherries on Sarah's cakes=5(2x+4)\n\nBut number of cherries is equal so:\n\n4(3x-5)=5(2x+4)\n\nPhew!! I think that works!! Lol!
902875
It sounds like random sampling... why?\nI am not sure what you mean, if it were simply, say, the first 100 names in the Manchester telephone system's number data base, then it would not be random because only a relatively few people whose last name began with an A would be included.\nNow, if all those numbers were mixed up in a hat, and the first 100 drawn were to be used, then the sample is random because anyone in the number data base could theoretically been chosen.\nLet's have a little fun with this. \nIt is impossible to sample, say all 110,000 chocolate bars made by a company in one month. However, quaity control wants to make sure each has a certain number of nuts in them\nSo as the bars are completed at end of the assembly line, once a week, a quality control person randomly grabs 100 bars and counts the nuts in it. Again, any of the 110,000 could have been chosen, but the person merely grabs 100 of them.\nThrough inferential statistics the company can estimate with high accuracy if all the candy bars produced have the right amount of nuts by merely counting nuts in 100 bars. The key: ALL bars had an equal chance of being selected; this is what made the sampling random.
624712
This problem is solvable.\nProbability of event A is written P(A).\n\nP(A)= Total # of outcomes that correspond to event A (divided by -or- over) Total # of outcomes in sample\n\nAnswer:\nP(A) = 12/20 = 3/5
913827
Let b = brown chocolate cakes\nLet w = white cakes\nLet c = cherry cakes\nLet a = angel food cakes\nLet p = pound cakes\n\nb = 3w\nw = c - 2\nc = 2a\na = p + 2\n\nAnd b + w + c + a + p = 62\n\nLet's put everything in terms of p:\na = p + 2\nc = 2a = 2(p + 2) = 2p + 4\nw = c - 2 = 2p + 4 - 2 = 2p + 2\nb = 3w = 3(2p + 2) = 6p + 6\n\nNow add them all together:\n6p + 6 + 2p + 2 + 2p + 4 + p + 2 + p = 62\n\nSimplify:\n12p + 14 = 62\n12p = 48\np = 4\n\nNow just solve for the rest:\na = p + 2 = 6\nc = 2p + 4 = 12\nw = 2p + 2 = 10\nb = 6p + 6 = 30\n\nSo 30 chocolate cakes, 10 white cakes, 12 cherry cakes, 6 angel food cakes and 4 pound cakes.
261662
coffee and chocolate.
1040231
Put chocolate on the socks in question
1106034
CHOCOLATE!!!!!!!!!!!!! and ice cream and chex mix\nYUUUUMMMM!
1147882
Life is like a box of chocolates....\n\nI eat the good ones (dark, hazelnut, caramel) then feed the rest to my family! ha ha!
997897
Chocolate and more chocolate.