Mathematical Question  (Probability)?
403576
In game one between A and B, the outcomes are:  A wins, B wins, or they draw, equaling 3 outcomes.\n\nIn 14 games, there are 3^14 outcomes.
631161
This is a binomial probability.\nP(success) = 1/5, P(failure) = 4/5.  There are 6 events here, so P(2 successs) = (1/5)(1/5)(4/5)(4/5)(4/5)(4/5).\nP(at least 2 successes) = P(2 successes) + P93 successes) + . . . . P(6 successes) or 1 - P(O successes) + P(1 success) + P(2 successes) which you already computed above.\nLet me know if you don't understand.
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
1169384
The previous responders are incorrect. \n\nLet the expected length of the game (in tosses) be L. We can classify each game into one of three starting sequences (in the following table P=probability,E=expectation value for total length of game):\n\nH --> P = 1/2 , E = L+1\nTH --> P = 1/4 , E = L+2\nTT --> P = 1/4 , E = 2\n\nThe first two cases are equivalent to starting the game over after 1 and 2 tosses respectively.\n\nSumming P*E for these three events gives L:\n\nL =(1/2)*(L+1) + (1/4)*(L+2) + (1/4)*(2)\n= (3/4)L + 3/2\n\n--> L = 6
366018
try this site\n\n\nhttp://www.futureaccountant.com/probability/
354101
This is easy, just say there are 100 shaft holders.  So of A, 50 of them are processed there, and 3% don't work, so 1.5 out of 50.  For B, it is 30 x 4%=1.2, and C is 20 x 5%=1.  \n\nSo out of a total 100, 3.7 don't work, the the probability that this one chosen at random wont work is 3.7%.  \n\nHopefully the way i calculated the answer helps you out in the future.
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.
220111
Greetings:\n\nHaving selected one of the four, the probability that person #2 will have a different date of birth is 364/365 (if we seek no repeats, #2 can have any 364 of 365 possible b.days).  Similarly, having selected two, the probability that #3 will have a unique d.o.b. is 363/365 (any 363 of 365 possible dates okay).  Finally, for #4, the probability is 362/365.  Hence by the multiplication property, the probability that all four will have different b.days is given by, (364/365)(363/365)(362/365) = (364)(363)(362)/(365)^3 = 0.984 to three places.\n\nRegards,\n\nNikkor180
872621
well in my skool we take it as {<and rule>} and {<or rule>example:there is 5 white marble and 5 black marbles in a bag...u pick 2 randomly without replacing them in the bag.....wat is the probability that u get 2 white marbles==>this is a combination 2 the way it is solved is multiplying the the 2 combined probabilities\nanswer:pr:(white AND white)=(5/10)*(4/9)\n\nnow the set is this way....wat is the probability that u get 2 marbles of different colors in the first 2 picks===>set rule\n\nanswer:there is 2 combinations in which u can get 2 different colors there are {white then black} OR {black then white} u find the 2 combinations then ADD them\npr(WandB OR BandW)={(5/10)*(5/9)}+{(5/10)*(5/9)}
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/
1097980
0.8037 or 80.37%\n\nThis is why:\n\nP(E)=1-P(not E); so the probability of at least 2 people having the same number is 1- the probability that each person has a unique number.\n\nThe probability that each person has a unique number:\n\nFirst person has 100 choices out of 100 numbers, so 100/100.\nSecond person can't choose the first person's number, so only 99 choices out of 100, so 99/100;\nThird only 98/100;\n. . .\n18th person, has (100-17)/100=84/100\n\nTherefore the probability that none of these people have the same number is (100/100)•(99/100)•(98/100)•(97/100)•. . .•(84/100) ≈19.63%\n\nSo the probability that at least two people have the same number is 100%-19.63%=80.37%.
261322
In the experiment of rolling two dice\n\nTotal number of possible choices = 36 ==> n=36\n\nLet "A" be the event of rolling the dice such that the sum on the dice is 7\n\nFor Event "A"\n\nNumber of favorable choices = 6  ==> m = 6\n[(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)]\n\nProbability of getting a sum of 7 on rolling the dice \n==> Probabilty of occurance of event "A" = m/n\n==> P(A) = 6/36\n==> P(A) = 1/6\n\nhttp://www.futureaccountant.com/\nhttp://www.schoolingkids.com/probability/
435037
try this site for help on probabiliity\n\nhttp://www.futureaccountant.com/probability/
1155743
Mutually exclusive - They both cannot occur at the same time.\nTherefore P(A and B) = 0\nP(A or B) = .5
694267
Probability choose red the first time: 5/8\nAFter choosing red, have 4R and 5W. Therefore, probability of choosing Red again is 4/9. So, probability of choosing red and then red, is 5/8 * 4/9 = 5/18\n\nIf choosing white the first time: 3/8 chance of choosing white first. Then there are 7R and 2W, so probability of choosing red is 7/9. Probability of choosing white then red is 3/8 * 7/9= 7/24.\n\nAdd these two to find answer for first part. 5/18 + 7/24 = .56944\n\nFor second part, keep the proabailty of finding red and then red, and find the probabilty of finding white then white. Probabilty of finding white first time is 3/8. Then have 7R and 2W, so probabilty of finding white now is 2/9. Multiply the two 3/8 * 2/9 = 1/12.\n\nAdd 5/18 + 1/12 = .36111 as the answer for the second part.
548980
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.
727752
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