a kitchen drawer contains 7 forks, 4 spoons, and 5 knives.  three are selected at random find each probability\n\n1. P(3 forks)\n\n2. P(2 forks, 1 Knife)\n\n3. P (3 spoons)\n\n4. P(1 fork, 1 knife, 1 spoon)
121373
For and event, odds are ususally expressed as the number of possible outcomes that qualify as a success compared to the total number of possible outcomes: 1 in 6, 3 in 10, etc.\n\nProbability is the same thing expressed as a fraction:\n1/6 = .66... 3/10 = .3\n\nWhen you select k objects from a set of n objects, the number of possible ways that you can do this is n!/(k!(n-k)!).\n\n\nIn all of your problems, your event is selecting a set of 3 objects from a set of 16. The number of possible outcomes of this is: \n16!/(3!13!) = 16x15x14/6 = 8x5x14 = 560.\n\nThe number of outcomes that are considered to be a success are:\n\n3 forks (pick 3 forks from the set of 7) = 7!/(3!4!) = 7x6x5/6 = 35\n\n2 forks 1 knife  (pick 2 forks from the set of sevan and 1 knife from the set o 5) = 7!/(2!5!) x 5 = 7x6/2 X 5 = 21x5 = 105\n\n3 spoons (Select 3 spoons from the set of 4) = 4!/(3!1!) = 4\n\n1 fork, 1 knife, 1 spoon  (Select 1 fork from the set of seven, 1 knife from the set of 5 and one spoon from the set of 4) = 7!/(1!6!) x 5!/(1!4!) x 4!/(1!3!) = 7 x 5 x 4 = 140\n\n1. P(3 forks) = 35/560\n2. P(2 forks, 1 Knife) = 105/560\n3. P (3 spoons) = 4/560\n4. P(1 fork, 1 knife, 1 spoon) = 140/560
299433
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
114320
your problem has been solved on this page...\n\nIts difficult to present the solution on yahoo answers.\n\ncheck this link\n\nhttp://www.futureaccountant.com/probability/queries/galaxy.php\n\n\nhttp://www.schoolingkids.com/
182848
I can't believe that this question is being asked in the sixth grade.  Maybe half of my college students could answer this.\n\nThe best way to answer this question is to make a table.  Make the rows the results of the first die and the columns the results of the second die.  Then put the product of the two roll inside of the table.\n            1   2   3   4   5   6\n          ------------------\n        1| 1   2   3   4   5   6\n        2| 2   4   6   8  10  12\n        3| 3   6   9  12  15  18\n        4| 4   8  12  16  20  24\n        5| 5  10  15  20  25  30\n        6| 6  12  18  24  30  36\n\n(Sorry I can't get the columns to even out!)\n\nNotice that there are 36 possible rolls of the two dice.  To get the two probabilities, just count the number of times you get the event in the table and divide by 36.\n\nProbability of an even product = 27/36\nProbability that the product is less than or equal to six = 14/36\n\nI see a lot of incorrect answers up there.  I teach statistics at the college level.  Trust me, this is right.
297736
Probability\n= P(boy and anyday) and P(boy and that same day)\n= (1/2)(1) * (1/2)(1/365)\n= (1/2)*(1/730)\n= 1/1460
38610
i don't believe the person above me got the second part right.  he is right about the beginning - the first probability is 2/5:\n\npossible numbers to draw: 2 4 6 11 13 15\n\nso you can choose any of these six out of the fifteen given (6/15 = 2/5).  the next number you draw must be also a "winning" number.  BUT, you have to have an even number less than 7 AND an odd number greater than 10.  this gives us two scenarios:\n\nIF the first number picked was even and <7, then the possible numbers you can pick from are ONLY: 11 13 15, since picking another even number <7 will not meet your requirements.  this gives us a probablity of 3/14, since there are 3 "winning" numbers out of the 14 numbers left.\n\n\nIF the first number picked was odd and >10, then the possible numbers you can pick from are ONLY: 2 4 6, since picking another odd number >10 will not meet your requirements.  this gives us a probablity of 3/14, since there are 3 "winning" numbers out of the 14 numbers left.\n\nSO, finally, suppose you go with the first scenario.  the probability would be: (6/15)*(3/14) = 3/35  going with the second scenario will give you the same thing, 3/35.  thus adding these two, since you can have either scenario occur and still meet your requirements, gives you: 3/35+3/35 = 6/35.\n\nAnd that would be the odds: 6/35.
272863
since it is "at least", use 1-(probability of no aces or no kings), so:\n\n1-((44/52)*(43/51))\n\nfor the second problem, I would use the idea of (probability of no orange + probability of one orange), so:\n\nprob of no orange:\n((5/8)*(4/7)*(3/6)*(2/5))\n\nprob of one orange, assuming you've learned combinations and permutations:\n(3C1*5C3)/8C4\n\nadd the two prob from above.\n\nhope this helps...probabilty can be quite ambiguous.
5936
There are 6 combinations\n3,5,7\n3,7,5\n5,3,7\n5,7,3\n7,5,3\n7,3,5
213154
P(A or B) = P(A) + P(B) - P(A and B)\n\nP(A) = probability of trying out for basketball\nP(B) = probability of trying out for baseball\n\nNow, P (A and B) = 0, since Jamie will not try out for basketball and baseball\n\nP(A or B) = 1 (or 100%), since Jamie will definitely try out for basketball or baseball\n\nSo, P(A) + P(B) = 100%\nP(B) is more than 42% or more (since you have to consider that Jamie will try out for other positions for baseball).\n\n100% - 42% = 58%\nP(A) has to be 58% or lesser. The only answer that is less than 58% is option A (40)
153517
This is a simple problem, so a straight-forward solution is to draw the set of all possibilities on a piece of paper and count them.  (And you should read this carefully, because it is the first answer so far to get all  parts correct.)\n\nProblem A)\n\nThere are 36 ways to roll 2 die:\n\n66 65 64 63 62 61\n56 55 54 53 52 51\n46 45 44 43 42 41\n36 35 34 33 32 31\n26 25 24 23 22 21\n16 15 14 13 12 11\n\nOf these, there are 11 ways to roll at least one 5 (Event A):\n\n$$ 65 $$ $$ $$ $$\n56 55 54 53 52 51\n$$ 45 $$ $$ $$ $$\n$$ 35 $$ $$ $$ $$\n$$ 25 $$ $$ $$ $$\n$$ 15 $$ $$ $$ $$\n\nSo, P(A) = 11/36\n\n\nProblem B)\n\nOf the 36 ways to roll 2 die, there are 26 ways for Event B to occur:\n\n66 65 64 63 62 61\n56 55 54 53 52 51\n46 45 44 43 42 $$\n36 35 34 33 $$ $$\n26 25 24 $$ $$ $$\n16 15 $$ $$ $$ $$\n\nSo, P(B) = 26/36 = 13/18\n\n\nProblem C)\n\nNotice that if Event A occurs, so does Event B.  Therefore, there are 11 ways for Event (A and B) to occur:\n\n$$ 65 $$ $$ $$ $$\n56 55 54 53 52 51\n$$ 45 $$ $$ $$ $$\n$$ 35 $$ $$ $$ $$\n$$ 25 $$ $$ $$ $$\n$$ 15 $$ $$ $$ $$\n\nSo, P(A and B) = P(A) = 11/36\n\n\nProblem D)\n\nNotice that if Event B does not occur, neither does Event B.  Therefore, there are 26 ways for Event (A or B) to occur:\n\n66 65 64 63 62 61\n56 55 54 53 52 51\n46 45 44 43 42 $$\n36 35 34 33 $$ $$\n26 25 24 $$ $$ $$\n16 15 $$ $$ $$ $$\n\nSo, P(A or B) = P(B) = 13/18\n\n\nProblem E)\n\nP(A|B) means the probability that Event A occurs given that Event B also occurs.  In this instance, there are only 26 ways to roll the dice so that B occurs:\n\n66 65 64 63 62 61\n56 55 54 53 52 51\n46 45 44 43 42 \n36 35 34 33 \n26 25 24 \n16 15 \n\nOf these 26 possibilities, there are 11 ways that Event A could occur:\n\n$$ 65 $$ $$ $$ $$\n56 55 54 53 52 51\n$$ 45 $$ $$ $$ \n$$ 35 $$ $$ \n$$ 25 $$ \n$$ 15 \n\nP(A|B) = 11/26\n\nAlternatively, you can directly use the definition of conditional probability, which is:\n\nP(A|B) = P(A and B)/P(B)\n\nSo, P(A|B) = (11/36)/(26/36) = 11/26
105970
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.
544553
The sample space consists of all pairs (x,y) with 1 <= x <= 6 and 1 <= y <= 7. There are 6 * 7 = 42 in elements and each is equally probable. You only need to count up the number of elements in each event (subset of the sample space) and divide by 42 to get the answers:\n\n(a) 1 (1,1) --> P = 1/42\n(b) 6 (2,x)'s +  5 (x,2)'s --> P = 11/42\n(c) 9 (even,even)'s --> P = 9/42\n(d) 12 (even,odd)'s + 9 (odd,even)'s --> P = 21/42 = 1/2
77842
In the experiment of choosing a committee of 6 from a group of 8 women and 7 men,\n\nTotal no. of possible choices \n= no. of ways in which the 6 members of the committee can be chosen from the total 15\n==> n = C(15,6)\n==> n = 15*14*13*12*11*10/6*5*4*3*2*1\n==> n = 7*13*11*5\n==> n = 5005\n\nLet "A" be the event of choosing the committee with 3 women and 3 men\n\nFor Event "A"\n\n---------- Men----Women---Total\nAvailable---8-------7------15\nTo Choose---3-------3-------6\nChoices----C(8,3)-C(7,3)--C(15,6)\n\nNo. of favorable choices\n= (no. of ways in which the 3 men can be chosen from the total 8) * (no. of ways in which the 3 wommen can be chosen from the total 7) \n==> m(a) = C(8,3) * C(7,3)\n==> m(a) = 8*7*6/3*2*1 * 7*6*5/3*2*1\n==> m(a) = 8*7 * 7*5\n==> m(a) = 1960\n\nProbabilty of choosing the committe with 3 women and 3 men \n==> Probability of occurance of event "A" = m(a)/n\n==> P(A) = 1960/5005\n==> P(A) = 392/1001\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/
413399
Binomial distribution, (a) p=0.5, n=12; (b) p=0.5, n=13.  You can do it!
525847
P(At least 2 orange) = P(2 orange or 3 orange)\n= P(2 orange) + P(3 orange) (These event are mutually exclusive, meaning that these events cannot occur at the same time.  This uses an addition rule.)\n= P(OON or ONO or NOO) + P(OOO) (OON means orange 1st and orange 2nd and non-orange 3rd.)\n= P(OON) + P(ONO) + P(NOO) + P(OOO) (also mutually exclusive)\n= (3/11)(2/10)(8/9) + (3/11)(8/10)(2/9) + (8/11)(3/10)(2/9)\n+ (3/11)(2/10)(1/9) (Since there are ands here, you would use the multiplication rule)\n\nThis gets you approximately 0.152 as the answer.\n\nThe other people are not answering correctly.  2/11 is the probability of getting an orange candy if you select one.  That is not what was asked.  Three are being selected, not one!  Jacob, just because you rounded correctly doesn't mean your answer is correct.
399137
1. If you don't care about color, then you just choose 5 from 13.\n\n13C5\n\n2.  For this one you just choose 2 gold fish from the 5 and 3 black fish from the 8.\n\n5C2 * 8C3
533490
a) \nThe probability Anna passes is 1/2\nThe probability Boris passes is 4/9\n\nThe probability that they both pass is the product of the two probabilities:\nBoth = 1/2 * 4/9 = 4/18\nReduce this to 2/9:\n\nProbability they both pass = 2/9\n\nb) \nFirst, a common mistake would be to assume that the chance that they both fail is 1 minus the probability they both pass, but it isn't.  Instead, first figure the probability that Danny passes.  This is just the reverse of the problem above.\n\nThe probability Chloe passes = 3/5\nThe probability Danny passes = d\nThe probability they both pass = 7/15\n\n3/5 * d = 7/15\nd = 7/15 / 3/5\nd = 7/15 * 5/3\nd = 35/45\nd = 7/9\n\nSo the probability that Danny passes is 7/9\n\nTo figure the chance they both fail, first figure their individual probabilities of failure.\nProbability that Chloe fails = 1 - 3/5 = 2/5\nProbability that Danny fails = 1 - 7/9 = 2/9\n\nProbability they both fail is the product of 2/5 and 2/9.\n\nChance they both fail is:\n2/5 (Chloe fails) x 2/9 (Danny fails) = 4/45