Probability Question... At a dinner party, 5 people are ...?
960573
the answers that the above people have given are wrong. the circular table does matter. let's say A, B, C, D and E are the people around the table. if they were in a line, ABCDE would not be the same arrangement as BCDEA. But in a circle, they are the same. It is unfortunate that I have forgotten what the formula is. search for it on google and you should get some answers. \n\nhope i helped.
733920
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.
599149
The probability of getting heads on one toss is 1/2 (or 50%). To figure out the probability of getting heads on 5 consecutive tosses you just multiply the probability of a single toss 5 times:\n\n1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/32\n\nThe probability of getting all tails is the same, for the same reasons: 1/32.\n\nTherefore, the probability of getting either all heads or all tails is simply 1/32 + 1/32 = 2/32 = 1/16
812841
Given 5 children there are 2^5 or 32 possible combinations of girls and boys.  Since there is only one combination with all boys that leaves 31 combinations containing at least 1 girl.  Therefore the probability is 31/32.
430637
First choose which number it is.  There are 6 numbers, so there are C(6,1) = 6 choices.\nThe probability of that number is 1/6\nSince there are three\n(1/6)(1/6)(1/6) = 1/216\nNow multiply this times 6\n6/216 = 1/36
206747
If, as you say, you're not concerned about the 6th ball:\n\nOn the first ball, there are 49 total balls, and 5 of them are the in the winning combination.\n\nOn the second, there are 48 balls, 4 winners.\n\nEtc.\n...\n\n(49/5) * (48/4) * (47/3) * (46/2) * (45/1) = 1,906,884 possible cominbations\n\nYou would need 1,906,884 tickets.
911469
I'll do some for you to show how they are done. In all cases probability P = #number of good throws/ # of all throws.\nAlso 1-P = #number of bad throws/ # of all throws\nOften you can arrive at a solution by considering the dices one by one.\n1) First dice can be everyting, second dice can be one of the five numbers not yet thrown, third dice can be one of the four numbers not yet thrown, etc. This leads to P = (6/6) * (5/6) * ... *(1/6) =720/46656 = 0.0154\n2) Divide the dice in three pairs, named A,B and C. The probability to throw a pair of two's with pair A is 1/36. Idem for B and C. But we have 6 posibilities to arrange the pairs ABC, ACB, ... And we have 6x5x4x3x2x1 possibilities to arrange the dice into the three pairs. Therefore P = 6 x 720 x (1/36)^3 = 0.0926\n3) First dice 1, second dice 1, third dice 1 gives a probability of (1/6)^3 = 1/216. But there are 20 possibilities to arrange the dice.\nP=20/216 = 0.0926
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!)
855670
Boy, this is tricky.  I think the probability is (8! * 41c2 * 10c2)/49!, where 41c2 = 41!/39!2! and 10c2 = 10!/8!2!  In this number, 8! represents the correct picks, 41c2 the incorrect picks, and 10c2 the number of ways to order the picks.\n\nFor exactly seven correct numbers in 10 picks , only a minor change is needed.  It becomes (7! * 42c3 * 10c3)/49!.  Follow this pattern for 6 of 10, 5 of 10, etc
438798
if 4/12 of 315.\n\nso divide 315 into 12 groups and the total number of people in each group r = 26.25 \n\nnow u take 4 groups of 26.25 members. that means u multiply 4*26.25 = 105.\n\nthen u take 105 and subtract it by 315 cuz there r 315 people in the book club right?\nso its 315- 105 = 210 females in the book club!
683930
Go to the table with 7 seats and 3 pizzas.\n\n5 people eating 2 pizzas mean that 2 pizzas get divided among 5 people, so each person gets 2/5 of a pizza, or 0.400.\n\n7 people eating 3 pizzas mean that 3 pizzas get divided among 7 people, so each person gets 3/7 of a pizza, or 0.429.\n\n12 people eating 5 pizzas mean that 5 pizzas get divided among 12 people, so each person gets 5/12 of a pizza, or 0.417.\n\nThe 7 seater table has the most pizza per person.
765309
It depends if one person can be in a room or if all 4 people can be in the same room.  If only one person can be in a room at any given time, then it would be \n4*3*2*1=24 different ways\n\nIf everyone can go in the same room, then there are 4^4=256 different ways.
904590
from 5 women 3 women can be chosen in 5C3 = (5*4*3)/(1*2*3) = 10 ways\n2 men can be chosen from 4 men in 4C2 or 6 ways\n\nso total number of ways = 10* 6 = 60
194085
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/
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.
898818
There are sixty-six possible outcomes in the first two scenarios, and fifty-five possible outcomes in the third (Mary does not win):\n\nWhat is the probability a married couple wins the prizes? \n9.09% (6/66 x 100)\n\nWhat is the probability one prize goes to a man and one to a woman? \n54.5% (36/66 x 100)\n\nIf Mary does not win a prize, then what is the probability that John will? \n18.2% (10/55 x 100)\n\n(or 0% if John never made it to the party!)
554382
Matthew rolled four dice. What is the probability of the total of the spots equaling 14? The answer is 1/6.\n\nJenna decided to repeat Matthew's experiment, except she rolled six dice. What is the probability of the total of the spots equaling 21? The answer is 1/6.\n\nLarry decided to repeat Jenna's experiment, but with ten dice this time. What is the probability of the total of the spots equaling 35? The answer is 1/6.\n\nGive a formula as to show when 2n dice are rolled, which total of dice will have a probability of 1/6 of being rolled.