If a teacher chooses 10 questions out of 20 for an exam and I know the answers to 15 of them, what is the probability that I will know exactly 8 of the answers of the 10 questions? \n\nI already know the answer, but I need an explanation of how that answer was derived, so please include your reasoning.
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The answer should be 64350/184756 or 34.8%\nIs that what you have?\n\nTo solve this problem you will need a function called "choose", often written as C.\n\n20C10 means "given 20 items, how many possible combinations of 10 are there?".  If there are 20 questions and the teacher chooses 10 of them, then there are 20C10 possible combinations of questions on the test.\n\n20C10 is solved by 20!/[(10!)(20-10)!]\n\nThe solution to the problem is (15C8)*(5C2)/(20C10).\n\nThere are (15C8)*(5C2) possible combinations of 8 (out of the 15) questions that you do know and 2 (out of the 5) questions that you don't know.\n\n"Choose" functions are very long calculations so you should not be expected to do them by hand.  On a TI graphing calculator, use the nCr option under the MATH/PRB menu.
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You need to specify a probability distribution for the number of incoming students. That would most likely be modeled by a Poisson distributed random variable (as quite often in Queuing Theory) with rate lambda>0. I will denote it by X and the number of teachers by n..\n\nThen the criterion for the number of teachers reads as follows:\n\nP(X>n*2.5)<=0.1\n1-P(X<=n*2.5)<=0.9\nP(X<=n*2.5)>=0.1.\nF(n*2.5)>=0.1.\n\n(where F is the cdf of the Poisson distribution)\n\nYou can solve it for n, or, for large n, you can also use a normal approximation by using the central limit theorem. Tell me if you're interested how this is done by using "Add details".
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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%
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each child can be a male or a female \ntherefore the total no of events possible is 2^6=64\n\nthe question says that you should have atleast 3 girls \ni.e there can be three, four ,five or all six girls .\n\nthe number of ways of choosing 3,4,5,6 girls from 6 children is the sum\n6C3 + 6C4 + 6C5 + 6C6 (here C stands for combinations )\n\ni,e\n\n(6*5*4)/(3*2*1) + (6*5*4*3)/(4*3*2*1) + (6*5*4*3*2)/(5*4*3*2*1) + 1\nthis reduces to \n20+15+6+1\n=42\n\ntherefore the probability is \n42/64
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I remember once having a teacher give us a copy of a multiple choice test, telling us that that would be the EXACT test.\n\nWhat did I do?  I took a Magic Rub eraser and systematically punched holes in it with my mechanical pencil.  Each row of pinpricks would represent an answer to a question.  1 hole for an A, 2 for a B, 3 for a C, and so forth.  If anyone else saw it, they wouldn't suspect anything at all because they wouldn't know the orientation of the rows. How'd I do on the test?  A+!\n\nFor you, I'd say this.  Have your group sit within visible distance of each other's hands.  Whatever signals you use, don't make noise of any sort.  Wave your pencil sideways "x" times to signal the question number.  Nod the pencil up and down "x" times to signal the answer (1 for a, 2 for b,c, or d).  Then have a "reset" signal in case you incorrectly signaled something, maybe scratch your head.
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You must remove one green chip, 4 red chips and 4 blue chips becos now you hav 21 chips- which is 9 greens , 6 reds and 6 blues, now the probability of picking out a red chip would be 6 out of 21 - simplify this to get 2 out of 7 which is what the question is asking you..... and the probability of picking out a blue chip would also be 6 out of 21......hope this has helped alot with your question.  :P\n~cheers~
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After the first card is drawn, the deck has 25 red cards and 26 black cards.   (The deck started out with 26 of each; I'm assuming the jokers were removed earlier.)   So when you draw the 2nd card, the probability of it being red is\n\n    25\n-----------\n  25+26\n\nor slightly less than half.   (It doesn't matter that the first card was a heart rather than a diamond, for this problem.  That's just an oblique way of saying it was a red card.)\n\nIs this a homework problem?
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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
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Here's a step by step explanation of how the probability is assessed.  Hope this will make things clearer for you.\n\nhttp://www.mste.uiuc.edu/java/java/birthday/explanation2.html
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There are 10 different digits\nso p(x)=1/10=0.1\nwhere p(x) means the probability of number 'x' being selected.\neg probability of the number 3 being selected,\np(3) = 0.1
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Suppose there are two bowls full of cookies. Bowl #1 has 10 chocolate chip cookies and 30 plain cookies, while bowl #2 has 20 of each. Fred picks a bowl at random, and then picks a cookie at random. We may assume there is no reason to believe Fred treats one bowl differently from another, likewise for the cookies. The cookie turns out to be a plain one. How probable is it that Fred picked it out of bowl #1?\n\nIntuitively, it seems clear that the answer should be more than a half, since there are more plain cookies in bowl #1. The precise answer is given by Bayes's theorem. But first, we can clarify the situation by rephrasing the question to "what’s the probability that Fred picked bowl #1, given that he has a plain cookie?” Thus, to relate to our previous explanation, the event A is that Fred picked bowl #1, and the event B is that Fred picked a plain cookie. To compute Pr(A|B), we first need to know:\n\nPr(A), or the probability that Fred picked bowl #1 regardless of any other information. Since Fred is treating both bowls equally, it is 0.5. \nPr(B), or the probability of getting a plain cookie regardless of any information on the bowls. In other words, this is the probability of getting a plain cookie from each of the bowls. It is computed as the sum of the probability of getting a plain cookie from a bowl multiplied by the probability of selecting this bowl. We know from the problem statement that the probability of getting a plain cookie from bowl #1 is 0.75, and the probability of getting one from bowl #2 is 0.5, and since Fred is treating both bowls equally the probability of selecting any one of them is 0.5. Thus, the probability of getting a plain cookie overall is 0.75×0.5 + 0.5×0.5 = 0.625. \nPr(B|A), or the probability of getting a plain cookie given that Fred has selected bowl #1. From the problem statement, we know this is 0.75, since 30 out of 40 cookies in bowl #1 are plain. \nGiven all this information, we can compute the probability of Fred having selected bowl #1 given that he got a plain cookie, as such:\n\n \nAs we expected, it is more than half.
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a) There are 16 combinations and 4 of them include exactly 3 heads (HHHT, HHTH, HTHH, THHH).  So the probability is 4/16 or 1/4.\n\nb) This question results in the same probability (4/16 = 1/4), except it asks for it as 'odds' which is the probability of success divided by the probability of failure.  That is 1 to 3 odds.\n\nc) The odds against heads are just the reverse.  Probability of failure divided by probability of success.  That is 3 to 1 odds against.
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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
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Answer 1: 14 questions answered right, 1 question partially right (there was no mention of not recieving points for partially right answers)\n\nAnswer 2: 22 people (though i think my math is off)
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1) this operation is made up of two steps, that means the probablility of the whole operation = the multiplication of the probablility of every step\n a) probaility of a 1 from the first box = 1/6, the other box = 1/7 , so P(Both will be 1's) = 1/7 * 1/6\n\nb) p(Exactly one of the will be a 2) = 1/6 * 6/7 + 1/7 * 5/6 \n\nc) P(Both will be even) = 3/6 * 3/7 \nd) P( Exactly one of them will be even) = 3/6 *4/7 + 3/6 * 3/7\n\ne) P(At least one of them will be even) = 1- both of the odd = 1- (3/6 * 4/7)\n\n>> as for the sample space it would take me forever to write it down! sorry\n\n2) Not sure about this one\n3) a) 5%\n    b) 16% >> dont know wut details to include its just a law
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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
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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.