If 4 letters are randomly put in 4 envelopes what is the probability that only 1 will be correct? Show formula
609118
You must be a math mathematician!!!!!!!!!  How did you come up with that answer????????????
815499
The correct answer is 15, but the proper formula is (6C4)/2\n\nBy the way, you can check this by figuring out how many groups of 2 you can have, since every group of four being chosen is the same as a group of two not being chosen.  For the first person, there are six people to choose from.  For the second, there are five.  However, since picking Person I and Person II is the same as picking Person II and Person I, you have only half as many possible combinations, or 15.
19818
try\nhttp://www.peterwebb.co.uk/probability.htm
547199
I'm assuming that you want the probability of drawing a heart followed by a queen.\nP(heart)=(13/52)=(1/4)\n\nAssume that the 1st card wasn't a queen, so (4/51) \n1st card is a queen, so (3/51)\nP(queen) = (4/51) or (3/51)\n\nor = add\n\nso P(queen) = 4/51 + 3/51 = 7/51\n\nP(heart) and P(queen) = 1/4 and 7/51\n\nand = multiply\n\nso P(heart/queen) = 1/4 x 7/51 = 7/204\n\nHope this helps\n\nTake care
621701
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
242656
This one is easy-assume a 50% chance each time, so its 1/2 X 1/2 X 1/2 X 1/2, so 1 chance in 16.  There is another way of figuring this, using combinations, like sn777 did, but sn777 made a logical error.  There are these possible combinations: 4 m, 3m 1F, 2m, 2F, 1m 3F, and 4F. So just looking at that, you would think its 25%. BUT, there is only one combination of 4M, while there are 4 combinations that make 3m 1F, and 6 combinations that make 2m and 2F, etc-so the total possible combinations of 4 children is 16, and 4M is only one of those.
67102
43 out of 400.\nThere is 120*120 pairs of minuites within which in which they can arrive resturent. And to be within 3 minuite, the diffrence of this of this  two pair have to be 3 or less. If the first one of this is between 4 to 117, there will be 7 possibility of  second one.\nAnd teher will be 6 for 3 and 118, 5 for 2 and 119,and 4 for 1 and 120.\nso total possibolity is 114*8+6*2+5*2+4*2 that is 828\nSo the probability is 828/14400=43/200.
771495
The easy part:\n\nExpected number correct: 5 = (1/4) * 20\n\nThe less-easy part:\n\nThe number of correct answers is binomially distributed, with p = 0.25 and n = 20.  You want to know about the probability that the number of success (labeled 'k') is 10 or greater.\n\nFor a more details on binomial calculations, see:\n\nhttp://www.itl.nist.gov/div898/handbook/eda/section3/eda366i.htm\nhttp://people.msoe.edu/~jorgensen/MA262_Binomial1.pdf\n\nSkip down to the middle of page 4 in this one:\nhttp://academic.csuohio.edu/holcombj/projects/Discrete_distributions.pdf
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.
147164
Ah, a variation on the Monty Hall problem:\n\nhttp://en.wikipedia.org/wiki/Monty_hall_problem\n\nIt is a probabilistic curiosity, this one. The highly unintuitive answer is that you actually **can** improve your chances by changing your mind.\n\nYour problem appears to only be different in that it starts with 5 boxes. The above link should give you sufficient information to solve the problem.
735176
Probability is the measure of how likely an event is. Is is given by the formula: \nP = (The Number Of Ways an Event Can Occur) divided by (The Total Number Of Possible Outcomes) \n\n1) To find the probability for a pen to be selected from the first box, you write the number of pens over the total number of items in the box. So: \n\n2 (number of pens) / 10 ( total items in the box) \n\nwhich reduces to 1/5. \n\n2) To find the probabilty for a marker to be selected from the second box, you do the same thing. \n2 (markers) / 5 ( total items in the box) which cannot be reduced. \n\nThe answers are 1/5 and 2/5. \nHope this helps ;-)
849665
The first two answers are the same answer ... 1/16 = 0.0625 ... and both are incorrect.  (no_life computed the odds that the first child only is the male ... but forgot about the cases where the second child is the only male, or the third child, or the fourth.  And Steph computed the odds that *all four* are boys ... in which case, yes it is 0.5*0.5*0.5*0.5.)\n\nThe correct answer is 1/4  = 0.25 = 25%.\n\nHow many possible combinations can you have?\n2x2x2x2 = 16\n\nHow many of them are "one boy + 3 girls"?\n4 ... the first child is a boy (and the rest girls), or the second child, or the third, or the fourth.\n\nSo the odds are 4/16 = 1/4 or "1 in 4" (or "4 to 1 against").
719566
The probability is the number of desired outcomes divided by the total number of outcomes.\n\nThere are 4 vowels out of a total of 11 letters.  Each letter has equal probability so the chance of a vowel is 4/11 (or 36.3636%)\n\nSimilarly there are 7 consonants out of a total of 11 letters.  The probability of a consonant is 7/11 or 64.6464%\n\nFortunately these add up to 11/11 or 100%, so that confirms our answers.\n\nIn summary:\nThe chance of drawing a vowel is 4/11 or 36.3636%
627413
OK.  Here are the possibilities.  It is important to keep the order of the dice in mind, so you can think of this as rolling a die 5 times.\n\nSSSSD\nSSSDS\nSSDSS\nSDSSS\nDSSSS\n\nNow each of these will have the same probability, so if I can find the probability of one of them, I can take that and multiply by 5 since there are five combinations.\n\nSSSSD can also occur in different ways.  If S=1, then D could be 2, 3, 4, 5, or 6.  So if S=1, then the probability of this combination is (1/6)^4(5/6).\n\nNow if S=2, then D can be 1, 3, 4, 5, or 6.  So again, (1/6)^4(5/6).\n\nSo hopefully you can see that there are 6 ways that you can do this.  So the probability of SSSSD is 6(1/6)^4(5/6).\n\nSo overall, the probability of getting four of a kind (without getting five of a kind) is\n5*6*(1/6)^4(5/6) = 150/6^5.\n\nSo it looks like your method does work.  Good job; this isn't an easy on.  There is always more than one way to look at probabilities.
342259
6! or 720 arrangements.\n\n2 * 4! or 48 with vowels at the ends, so it's a 1/15 chance.\n\nEdit: Yeah, I messed up.  Forgot the *2.
524692
Its the product of the probabilities of selecting each letter in order.\n\nProb(M) * Prob(a) * Prob(t) * Prob(h)\n2/11 * 2/11 * 2/11 * 1/11 = 8/14641\n\nIts really unlikely, about 5 in 10,000.
801480
4 is the answer I get