A jar contains three green marbles and five white marbles. Suppose you choose a marble at random, and do not replace it. Then you choose a second marble. Find the probability of each event.\n\n1. You select a green marble then a white marble.\n\n2. Both of the marbles you select are green.
201539
Okay...this may get a little confusing. :) \n\nThe first thing that you need to do is figure out whether the two event listed in each number are independent of each other or dependent.  In this case, the events are dependent.  For example, if you have these 8 marbles in a bag and you pick a green marble out of it...without replacement you would just set that green marble aside.  Now the chances of getting a green or white marble are different because you have changed the total number, does that make sense?\n\nNow, because the events (let's just work with number 1 for example but the same thing can be done to number 2) are dependent, you must use the multiplication rule.  In math terms this says that "The chance that two things both happen equals the chance that the first happens multiplied by the chance that the second happens pretending the first has happened".  In simpler terms this just means that you take the chance that the first will happen and multiply it by the chance that the second thing will happen.  \n\nSo in number 1, the first one you select is a green marble and the second one is a white marble.  The chances of getting the green marble are 3/8 because there are 3 green marbles and 8 marbles all together.  The chances of getting the white marble second are 5/7.  There are 5 white marbles in the bag, but because you have already taken the green marble out you have changed the total number of marbles - hence the 8 in the denominator changes to a 7...does that make sense?\n\nSo your answer to the first one would be (3/8 x 5/7).  I'm not for sure what the numerical answer to that would be, but you can do that calculation on your own! :) \n\nThe second problem can be done the same way, except instead of having a white marble second you would have another green marble.\n\nI really hope that this helps!! :) :) Good luck! :D
427802
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.
114576
This isn't probability - it's permutations.  Often times teachers will teach permutations during probability units because you use the one in the other.  \n\nAnyway, the best solution is to make an organized list.  For example, if you choose color one as the background, then how many stripe combos can you have?  This ends up being a simpler problem.  If you can keep making these problems simpler and simpler, get the answers then combine them all together, it will help you solve the hard ones.
504113
When you are computing the probability of a given event such as 2 heads, 2 tails, or one of each, you are essentially counting the number of ways a given event can occur and dividing this number by the number of all possible outcomes.\n\nIf you toss 2 coins once ( or one coin twice) there are 4 possible outcomes:\n\nHH = both heads\nHT = 1st heads, 2nd tails\nTH = 1st tails, 2nd heads\nTT = both tails\n\nAssuming the coins are "fair" each outcome is equally likely.\n\nThere is one way in four that two heads (or 2 tails) can happen so the probability for each of these outcomes is 1/4\nBut there are two ways you can have one head and one tail and so the probability of this happening is 2/4 =1/2. The sum of all the probabilities of the distinct outcomes will always equal 1. \n1/4 +1/4 + 1/2 =1\n\nHow about if you have 3 tosses? The 8 possible outcomes are as follows:\n\nHHH \nHHT\nHTH\nTHH\nHTT\nTHT\nTTH\nTTT\nWe see there is only one way to have 3 heads (or 3 tails) and so the probability of 3 heads (or 3 tails) = 1/8\nHowever there are 3 ways you can have 2 heads and one tail, so the probability of this happening is 3/8. And there are also 3 ways you can have one head and 2 tails, so 3/8 is the probability for this too.
114495
Assuming that all the gumballs are unique, then there are 8*5 = 40 different ways to draw two gumballs at random, one from each bag.\n\nJust to clarify, though - the concept illustrated here is NOT probability, but an illustration of the fundamental counting principle. Probability can be derived from this, but it's not the main thing illustrated here.
538014
1) From the data we know that the probability of a box having at least 1 defective bolt is 1/10. Therefore the probability of the box having no defective bolts is 9/10. To get 3 boxes that are good, your probability is (9/10)*(9/10)*(9/10) = 72.9%.\n\n2) As males and females make up half the population each, the probability that someone chosen at random is color blind is \nprobability(person=male)*probability(person=color blind) + probability(person = female)*probability(person=color blind) = 2.5% when you plug in the values.
537868
A.  The probability that it will BE solved is 1 - prob that it WON'T be solved.  That is 1.00 - (1.00 - .015 [Tom doesn't]) * (1.00 - 0.50 [Alice doesn't]).  \n\n = 1.00 - (.85) * (.50).  = 1.000 - .425 = .575\n\nB.  If Tom solves it then it is 150 / 575 = 6/23
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.
36137
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
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/
294273
11\nr,b.g.o\nr b g y\no y \n the next marble will make a set no matter the color\nyou could get a set before this if you draw well\nbut to be sure to get a set you need to draw 11
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.
386030
In total there are C(5,2) = 5!/2!3! = 10 ways to choose two officers among 5 people.\n\n1) There are C(3,2) = 3 ways to choose two women (among the three) so the probability is 3/10 = 0.3.\n\n2) There is only C(2,2) = one way to choose two men, so the probability of having two men as officers is 1/10.  Threrefore 0.9 for having at least one woman.\n\n3) This derives from the previous two answers.  There are 9 ways of having at least one woman officer, and 3 ways of having two.  So, 3/9 = 0.333.\n\n4) Knowing that Beth is an officer, this leaves one person to choose among two men and two women for the other position, so it's 2/4 or 0.5.
334586
As it is certain that the first one is red and the second one is blue, we can ignore the fact that they were even mentioned. (It would be different if you phrased it as find the probability of the first ball being red, the second being blue and the third being yellow...)\n\nP(Yellow) = 1/5
453882
The answer to this question is (b) 2/7.\n\nWhen removing two dogs there are four possible scenarios:\n\n(1) Remove poodle, then remove terrier.\n(2) Remove poodle, then remove another poodle.\n(3) Remove terrier, then remove another terrier.\n(4) Remove terrier, then remove poodle.\n\nIn each case, take the probability of the first even happening and multiply it by the probability of the second even happening. For instance:\n\nProbability that a poodle is first selected = 5(# of poodles originally) / 7(total # of dogs).\nProbability that a terrier is then selected = 2(# of terriers left) / 6(total # of dogs after a poodle is removed).\nMultiply 5/7 and 2/6 to get 10/42.\n\nRepeat this with each of the remaining three cases to get\n(1) 10/42\n(2) 20/42\n(3) 2/42\n(4) 10/42\n\n(Note that these add up to 1, which means that we have all of the possible cases)\n\nNow, just add the probabilities of the two cases with a terrier being selected second to get:\n\n10/42(first case) + 2/42(third case) = 12/42\n\nReduce 12/42 down to 2/7 to get your answer.\n\nI hope this helps, good luck with future probability problems!
303664
a. first draw doesnt matter since it's replaced; you're back at 3 in 10 chance\n\nb. two ways this could happen: first, 3/10 chance first ball is red, leaving you 2/9 chance then of second ball being red on second draw.  second, 7/10 chance you draw a white ball, leaving you with 3/9 chance of second ball being red.  (3/10*2/9)+(7/10*3/9) which leaves you with 3/10 chance. again.\n\n\nEDIT:\n\nsurprises me how many people can't do math.\n\nfor A), the second draw is completely independent from the first.  Doesn't matter if I took a dump in the bowl in the first draw; as long i fished it out, you're left again with the same start on the second draw-3 reds and 7 whites.  your probability doesn't change because of what happened in the first draw as the conditions are RESET.\n\nfor B), let's run through allll possible scenarios\nNumber the Balls\n\n1 - red\n2 - red\n3 - red\n4 - white\n5 - white\n6 - white\n7 - white\n8 - white\n9 - white\n10- white\n\nPossible draws (X denotes red second ball outcome)\n1,2 - X\n1,3 - X\n1,4\n1,5\n1,6\n1,7\n1,8\n1,9\n1,10\n(2 fav outcomes if ball 1 is drawn first)\n\n2,1 - X\n2,3 - X\n2,4 \n...\n2,10 \n(2 fav outcomes if ball 1 is drawn second)\n\n3,1 - X\n3,2 - X\n3,4 \n...\n3,10 \n(2 fav outcomes if ball 1 is drawn second)\n\n4,1 - X\n4,2 - X\n4,3 - X\n4,5 \n...\n4,10\n(3 fav outcomes if ball 4 is drawn first)\n\n...\n\n10,1 - X\n10,2 - X\n10,3 - X\n...\n10,8\n10,9\n(3 fav outcomes if balls 5-10 are drawn first)\n\n\nthere are 90 outcomes\nof which there are 2+2+2+3+3+3+3+3+3+3 favorable outcomes = 27.\n\n27/90 = 3/10\n\n\nSecond edit: guy below, 2*3 + 7*3 = 6 + 21 = 27, not 28
390713
CASE 1:\nThe probability that 2 green balls are selected from box A:\n1/2 (chance of picking box A)\nChance of picking first ball green (3/6)\nChance of picking second ball green (2/5)\n\nSo the combined odds are: 1/2 x 3/6 x 2/5 = 1/4 x 2/5 = 1/10\n\nCASE 2:\nThe probability that 2 red balls are selected from box B:\n1/2 (chance of picking box B)\nChance of picking first ball red (4/9)\nChance of picking second ball red (3/8)\n\nSo the combined odds are: 1/2 x 4/9 x 3/8 = 12/72 = 1/6