diet program varies between 5 Ibs & 10 Ibs. uniform distribution Find probability that more than 9Ibs are lost
95999
Don't you think you could have written this more clearly? \n\nIf there is a uniform distribution of the weight w being between 5 and 10, then each of these probabilities is equal:\n\nP5_6=P{5<=w<6}\nP6_7=P{6<= w<7}\nP7_8=P{7<=w<8}\nP8_9=P{8<=w<9}\nP9_10=P{9<=w<10} \n\nThe sum of these is 1, so \n\n1=P5_6+P6_7+P7_8+P8_9+P9_10, and\nP5_6=P6_7=P7_8-P8_9=P9_10\n\nso 1=5xP9_10, so P9_10=0.2\n\nThe probability is 0.2.
450313
10 percoset should do the trick\nThis is a joke....just in case you are dumb enough to try it.lol
165662
HOW MANY numbers do you have their  we have 6 and 2 supps...\nsoooo.. 9.11.27.35.36.44. supps. 10.23.\nHAVE A GO WITH THEM ONES  ...\ni hope you win good luck.
1171325
This is a binomial problem.  You have 20 trials, the probability of success is 0.35, and you want 10 successes.\n\n(20!/10!10!)(0.35^10)(0.65^10) = 0.069.\n\nI don't get the same answer, but this is correct.\n\nedit: I am aware of that.  That is what I did.\n\n(20C10)(0.35^10)*\n(1-0.35)^(20-10)\n\nIf you haven't noticed yet, I'm an instructor in statistics.\n\nedit: reetz, you have been reported again.\n\nedit: The probability that 10 or more would die does turn out to be around 0.12.  I used Excel to get that by typing =BINOMDIST(9,20,0.35,TRUE) and subtracting the answer from 1.  You can also get it by computing each binomial probability from 10 up to 20 and adding them together.\n\nIf the question is "that 10 or more would die", then that should have been stated in the problem.  When a question asks "that 10 sheep will die", that is interpretted as exactly 10 will die.  I can't help the grammar of the question.
938617
You can read off the outcomes from Pascal's Triangle. For 5 tosses consult the fifth row\n\n........1  1\n.......1  2  1\n......1  3  3  1\n.....1  4  6  4  1\n....1 5 10 10 5 1\n\nThere is \n1 chance in 32 of 5 heads and 0 tails\n5 chances in 32 of 4 heads and 1 tail\n10 chances in 32 of 3 heads and 2 tails\n10 chances in 32 of 2 heads and 3 tails\n5 chances in 32 of 1 head and 4 tails\n1 chance in 32 of 0 heads and 5 tails
701818
Gosh.   I have no idea how to do it in excel.... \n\nbut for a normal distribution with an error function that always returns 1 (no error)  then the percentage of the population between -n standard deviations and +n standard deviations is n/(sqrt(2) \n\nSo, if you want there to be 90% of the population between -n and +n so that 5% is below -n and 5% is above +n then \n\n0.9 = n/sqrt(2)\n0.9 * sqrt(2) = n\n\n1.27279221 = n \n\nso,  now we take 1.27279221 time the standard deviation of 0.87 and then subtract it from 10.42 \n\nUnderweight = 10.42 - (1.27279921 * 0.87)\n\n= 9.31266469\n\nand then you do the same thing for 80% (except you add) to get the weight at 10% over....
753276
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".
629177
P(selecting a person at random and he has the symptom and is detected by the procedure)\n\n= P(patient having it) and P(detecting it using the procedure)\n\n= (0.5/100)*(98/100)\n= 0.0049\n\nBut if the person is not randomly selected and you are sure s/he has the problem and its just a matter of detecting it with the procedure, then Probability = (98/100) = 0.98
220083
I am pondering this question a bit. I might have a better answer for you, but here is a start:\n- - - - - - - - - - - -  - \n\nYour data is not exactly amenable to standard quantitative analysis. You can't really form an average, for example. \n\nYou can find the mode. The most common answer is, diet. So the mode of your data is "diet". \n\nThe data categories, as they stand, are not represented by ordinal numbers. So it is a bit hard to even describe the distribution, let alone the dispersion of the distribution. What you could do, is say that 35% preferred diet, and 65% did not. Then you could try to estimate the error bounds on these figures, given some assumptions. Or do a robust estimate using a jackknife or bootstrap.\n\n- - - - - -  - - \n\nHere is another idea. You might want to know the average number of people that are in a given category. That is, you might want to know how many people like a given type of soda. You have 5 categories, so the average is \n\n(20)/5=4\n\nThere are an average of 4 people in each category.\n\nNow you can find the standard deviation.\n\nFirst find an estimate of the variance V using the standard formula:\n\nV=(1/4)*[(1-4)^2 + (3-4)^2 +(3-4)^2 +(7-4)^2 +(6-4)^2 ]\nV=(1/4)*[9 + 1 + 1 + 9 +4 ]\nV=(1/4)*[24]\nV=6\n\nThen take the square root\n\nS=sqrt(V)= sqrt(6)\nS=sqrt(6) = 2.44948974\n\nThe average soft drink is liked by 4 people, but there is a standard deviation of about 2.4.\n\nSo one can say a soft drink is liked by 4 += 2.4 people (if we set the error bounds to be 1 standard deviation, but some people set it at 2).\n\nYou could even convert this to percentages. Then you could say that a soft drink is liked by 20% of the people, +- 12%.\n\nThe standard deviation assumes that the distribution is normal, that is a bell-shaped curve (aka Gaussian). However, it probably is not. A robust estimate of the dispersion of the dispersion can be formed just resampling methods, like the bootstrap or the jackknife. I have placed a link to these if you are interested, but I think it might be too advanced for your needs.\n\nAnother thing to do is to look at the median, and the median average deviation, or the median average deviation away from the median. A reference discusses these and other measures of dispersion if you are interested.\n\nI still feel that one might be able to do more with this. I will think some more and if I have any good ideas I will post them. I am not a statistician, so it is likely that one of them might have a much better suggestion right away.
91512
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
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!)
763823
Who are you? \n\nWe cannot adequately answer this question without knowing a "pre-test" probability. That is,\n\nAre you young and athletic, or elderly with arthritis? \n\nBut theoretically, the probability of an event = the pretest probability times the probability associated with the event itself.\n\nThis is very confusing. So here's another way to look at this problem:\n\nThe probability of ALL equals the probability of getting up from the chair TIMES the probability of tripping down on the table TIMES the probability of falling on the ground, etc.\n\npA = pE1 * pE2 * pE3 etc.\n\nwhere pA = probability of A and pE1 = probability of event 1.\n\nSay the probability of gettin up from the chair is 100%, the probability of tripping down on the table is 1%, the probability of falling on the ground is 0.1%, the probability of swallowing your tongue is 1%, and finally the probability of suffocating is 50%. \n\npA = 1 * 0.01 * 0.001 * 0.01 * 0.5 = 5 EXP (-8) = 5 chances out of 100,000,000 or more simply, 50 chances in a billion.\n\nThis is assuming that the probability of each event is independent of each other event. But.... these events are NOT independent; each subsequent event depends on the previous event. \n\nThe probability that ALL events will happen =\n\n[ Probabilty of E1 ] * [ Probability of E2 given E1 has occurred ] * [ Probability of E3 given both E1 and E2 have occurred ] etc. \n\nIn real life, what most likely happens is someone will have a heart attack or diabetic attack, and THEN they will get up from the chair, feel dizzy and faint, falling on the table then to the ground, perhaps causing a seizure causing the person to swallow their tongue and suffocate.\n\nSo, who are you ?  Do you smoke ? Do you have high cholesterol? Are you young or old?
249369
Is the term you are looking for "probability"?\n\nI hope this helps.
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%)
484411
49/99\n\nProof:\n\n1) Probability that first passenger (P1) sits in last passengers (P100) seat (S100) is 1/99\n2) probability that P1 sits in S99 , and P99 sits in S100 is (1/99) * (1/2)\n3.1) probability that P1 sits in S98, and P98 sits in S100 is (1/99) * (1/3)\n3.2) probability that P1 -> S98, P98->S99, P99->S100 is (1/99) * (1/3) * (1/2)\n4.1) P1->S97, P97->S100 = (1/99)*(1/4)\n4.2) P1->S97, P97->S99, P99->S100 = (1/99)*(1/4)*(1/2)\n4.3) P1->S97, P97->S98, P98->S100 =  (1/99)*(1/4)*(1/3)\n4.4) P1->S97, P97->S98, P98->S99, P99->S100 = (1/99)*(1/4)*(1/3)*(1/2)\n5.1) P1->S96, P96->S100 = (1/99)*(1/5)\n5.2) P1->S96, P96->S99, P99->S100 = (1/99)*(1/5)*(1/2)\n5.3) P1->S96, P96->S98, P98->S100 =  (1/99)*(1/5)*(1/3)\n5.4) P1->S96, P96->S98, P98->S99, P99->S100 = (1/99)*(1/5)*(1/3)*(1/2)\n5.5) P1->S96, P96->S97, P97->S100 = (1/99)*(1/5)*(1/4)\n5.6) P1->S96, P96->S97, P97->S99, P99->S100 = (1/99)*(1/5)*(1/4)*(1/2)\n5.7) P1->S96, P96->S97, P97->S98, P98->S100 =  (1/99)*(1/5)*(1/4)*(1/3)\n5.8) P1->S96, P96->S97, P97->S98, P98->S99, P99->S100 = (1/99)*(1/5)*(1/4)*(1/3)*(1/2)\n\n1) (1/99)\n2) (1/99)(1/2)\n3) (1/99)(1/3)(1 + ½)\n4) (1/99)(1/4)(1 + 1/3)(1 + ½)\n5) (1/99)(1/5)(1 + ¼)(1 + 1/3)(1 + ½)\n\ni) (1/99)(1/i)product(j=2 to (i-1),(1+1/j))\n\n99) (1/99)(1/99)product(j=2 to 98,(1+1/j))\n\nGrand Sum = (1/99)*(1 + ½ +sum(i=3 to 99,(1/i)*product(j=2 to (i-1),(1+1/j))))\n=(1/99)*(1 + ½ + sum(i=3 to 99,(1/2)))\n=(1/99)*(1 + 98/2)\n= 50/99\n\nProbability that last passenger gets bumped = 50/99\n\nHence Probability that last passenger does NOT get bumped = 1 – 50/99 = 49/99\nA bit less than 50%\n\nWould be 1/2 like others suggest if first passenger was not constrained to sit in the wrong seat as given in the problem statement.
107013
Maybe there's a trick to this, but if the probability is\n5%, then if (for example) you opened 100 boxes, you would\nhave "gotten something" 5 times (ie. 5% of 100 = 5).\n\nThat seems like you'd get something, on average, once\nevery 20 boxes.
366018
try this site\n\n\nhttp://www.futureaccountant.com/probability/