A jar contains 50 marblesthat are the same size.There are 15 blue,8 yellow,and 27 red marbles find probability
485997
Blue: 15/50 (simplified to 3 out of 10)\nyellow: 8/50 (simplified to 4/25)\nRed: 27/50
279356
This is always so much fun, isn't it.  This is probably why the previous answerer is now looking for satisfying employment in the food-service sector.\n\nBy the way, you forgot to mention if they have to specifically be in that order or in any order, though.  It's a very different answer.\n\nOkay, what's the probability of drawing one green marble each time you draw, if you put the marbles back each time?  It's the partial over the sum, or in other words 4 out of 6, or 2/3.  Similarly the probability of drawing a red marble is 2 of 6, or 1/3.\n\nSo if you specifically want the order to be green, green, red, then you just multiply the probabilites each time.  2/3 * 2/3 * 1/3 is  4/27.\n\nHowever, suppose it can be in any order?  Here's where it gets tricky.  First we figure how many ways there are to draw two greens and a red -- we can draw (g,g,r) or (g,r,g) or (r,g,g) -- so three ways.  The probability is the same for each time so we add them together, essentially 3 * 4/27 is 12/27, or a little less than 50%.\n\nPart b) is harder, since the probabilities change with each draw.  The first time we draw it's a 2/3 chance of getting green, but after that there are only 3 green marbles out of 5, so it's now 3/5 instead of 4/6.  So if we want to draw two green and one red IN THAT ORDER it's 2/3 * 3/5 * 1/2 = 1/5.\n\nIf you don't care the order, then figure out each of the ways to draw 2 greens and one red (same as above) and add the probabilities from each time.  I'll leave it to you for the answer.\n\nActually there are very few interesting people in the world who like statistics.  But even the boring ones can usually find very good jobs.  Also it's been a while since my stats classes so if I'm wrong, please don't come after me with a baseball bat, thanks  :)
815169
I am assuming that you are drawing without replacing the marbles.\n\nThe probability of the first marble being red is 7/10.\n\nThe probability of the second marble being red is 6/9 = 2/3\n\nThe probability of the third marble being white is 3/8\n\nThe probability of all three events happening in that order is:\n\n7/10 * 2/3 * 3/8 = 7/40. This is 17.5%, or slightly less than 1 in 5.
257322
It's very simple. Every taster has to choose one out of 5 cups, so on every round the probability of a single taster choosing A is 1/5.\nSince the event is repeated 4 times independently, the overal probability will be 1/5 of 1/5 of 1/5 of 1/5.\nThat is, (1/5)^4 = 1/625.
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
1159931
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.
405508
5 marbles will make a row 3 inches long. (3/5*5=3), so double that. 10 marbles should make a row 6 inches long.
648450
First get it straight. Algebra and probability are 2 very different things! Now:\n\n1. There are 12 marbels, so,\na) purple then red = (1/12)*(3/12)=1/48\nb) red then green = (3/12)*(3/12) = 1/16\nc) 2 greens = (3/12)*(3/12) = 1/16\nd) 2 reds = (3/12)*(3/12) = 1/16\ne) purple then green = (1/12)*(3/12)=1/48\nf) red then yellow = (3/12)*(5/12) = 5/48\n\n2. There are 9 total marbles, so:\na) purple then white = (3/9)*(1/8) = 1/24\nb) white then green = ((1/9)*(1/8) = 1/72\nc) 2 purples = (3/9)*(2/8) = 1/12\nd) 2 yellow = (4/9)*(3/8) = 1/6\ne) yellow then purple = (4/9)*(3/8) = 1/6\nf)yellow then purple = (4/9)*(3/8) = 1/6\ng) green then white = (1/9)*(1/8) = 1/72\n\n13. Red card is 1/2 and club is 1/4, add them = 3/4\n14. Dimond is 1/4 ans spade is 1/4, add them = 1/2\n15. face card is 12/52 and a 2 is 4/52, add them = 16/52\n16. A 7 is 4/52 and a 9 is 4/52, add them = 8/52\n17. Red card is 26/52 and a king of spades is 1/52 = 27/52\n18. Heart is 13/52 and quuen od diamonds is 1/52 = 14/52
351129
Take the number of red blocks and divide by the total number of blocks.\nnumber of red = 18\nTotal number of blocks = 27+18+12+43 = 100\n18/100\n= 9/50
263161
This is a good probability problem -- I LOVE probability problems.  This is really asking 4 different questions.  First, what are the chances of the first pull being black.  (9 total socks, 5 are black ...).  Then what are the chances of the second one being blue (Keep in mind that now you only have EIGHT socks in the drawer, unless you put the last one back - so 1 blue and eight total socks).  Then chances of a brown sock with only 7 socks left (3 brown, seven total socks).  \n\nWhen you get all those probabilities, in order to see what the chances of this AND this AND this happening, you have to multiply the fractions (if it was "What are the chances of this OR this happening, you'd add them).\n\nGood luck.  This stuff is FUN!  (Especially if you ever want to go to VEGAS!)
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%)
568866
is there the same probability it will go into each bin?\n\nthen 350/9=38.8888 so about 39 in each
1082436
A) 2.  Since the probability of drawing a blue marble is 1/3, there must be 4 of them in the bag (4/12 = 1/3).  They also tell you that the probability of a blue or red is 1/2 = 6/12.  So 6 marbles are blue or red.  I know that 4 of them are blue.  So 2 of them must be red.\n\nB)  It says that the total group was 160.  So there must have been 160 people surveyed.\n\nC)  I don't get this part, though.  If 12 said they preferred vanilla and there were 160 people total, then 12/160 = 7.5% prefer vanilla.  But that isn't an option.  You must have copied something wrong somewhere.
714555
Odds of first marble being red is 4/10 or 2/5.\nOdds of second marble being red after first is selected is 3/9 or 1/3.\nOdds of both being red is 2/5 * 1/3 = 2/15.
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
1076427
Probability of 10 heads, 1/2^10 = 1/1024\nProbability of 9 heads    10/2^10 = 10/1024\nProbability of 8 heads    45/2^10 = 45/1024\n\nProbablity of at least 8 heads = \nProbablity of 10 heads +  Probability of 9 heads + Probability of 8 heads  = 56/1024 = 7/128.
1058390
Every week contains six days.\n\nSince six is less than seven, any time that something has seven, it has to also contain six.\n\nIf there are seven marbles in a box - you can also reach in and only pull out six.