What is the probability of the last passenger finding his own seat free?
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.
1119536
1) If no one is injured, then no subs ever get to play.  The probability is (.9)^11 = .3138106 or about 31.38%.\n\n2) (.9)^9 * (.1)^2 = .003874205 or about 0.39%\n\n3) "At least" is best calculated as the complement of "no substitutes were needed."  In the first calculation we determined the probability of not needing any substitutes.  So the probability of needing at least one sub would be .6861894 or about 68.62%\n\n4) The trick here is that the two injured could both be players, or one could be a player and the other could be the sub.  (They could not both be subs, however.)  The calculation in #2 (above) is part of this--the probability that two of the players were injured.  For the other part we need to know the probability that one player is injured (while 10 are not) and that the sub is also injured.  .1 * (.9)^10 * .15 = .0052302.  So the overall result for either of these outcomes (add this result with the result from #2) is 0.00910438 or about 0.91%.
86663
a win win situation if she leaves right after
98691
Most likely, yes.  Unless you bought the exact same kind of ticket from the exact same role before them in the exact same store, it wouldn't throw it off at all.  The odds of one person buying a ticket and throwing off who won the lottery, let alone winning it themselves are very very slim.
689280
You have to go to all the Airlines' office and check the menifest of the passengers, who boarded the flights. You should also tell them, when you suspect your wife to have left the country. It is a big job.
204526
What I usually do is time myself when picking someone up.  If they say that they land at 1:15 then I'll make sure I am there at 1:30.  The only reason I do that is because you have to keep driving around and you can't stop to wait so if you do that then hopefully that person will be out and waiting on you.  Baggage claim on the lower end is where you pick the person up.  Also look at the signs to point you in the direction of the airline that person used.  Its very simple!
777500
1) \n    a) The probability that the first one is not right is 7/8, the next is\n6/7 since there are 6 chances of getting the wrong one out of the remaining 7. The next is 5/6, then 4/5, 3/4,2/3,1/2. The probability of all these events happening is\n 7!/8! = 1/8\nMy colleagues who answered prior to my answer fail to\nrealize the difference between dependent and independent\ndistribution. In this case we have dependent distribution since each time a letter and envelope is matched we have 1 less\nof each to choose from.\n\n\nb) The probability that at least one envelope right is the additive\ninverse of the probabilty of no envelope getting it right based on\nall probabilities which 1. So 1 - 1/8 = 7/8.\n\n2)  The probability that no 2 have the same birthday is  computed like this. Based on a 365 day year just to make it reasonable. \n  First person - any day. \n Second - 364/365 'cuz 1 day is eliminated by the first one\nThird -363/365   'cuz 2 days are eliminated by the first two\n etc.\n The probability of no two the same is therefore\n (364)(363)(362)...(346)/365^19 = .588
380935
* How many people are there?\nA> Total seating capacity of the cars is 3200x4 + 4000x6 = 36800.\nIf, by (80% of the cars are full" you mean 80% of all cars, as an average, then total number of people = 80% of 36800 = 29440.\n\n* How many 4 seaters are there?\nA> 3200 (as mentioned in the question itself)\n\n* What is the combined driving time of all the cars?\nA> Total time is as follows:\n38% of all people live to 53 => 11187.2 live to 53 => lifetime = 11187.2x53 = 592921.6 years\nSimilarly, the numbers in the other cases are 31795.2, 194304, 296755.2 & 503424 years respectively. Adding, we get 1619200 years.\n\n* How many people are there of each age group?\nA> 53 : 38% of 29440 = 11187 and so on.
495404
keep your hands and feet inside the ride at all times... until the ride comes to a complete stop...\nthank you; come again!\n\nyep 2points/score!\n\ncall sunshine
271358
Probabilities must add to 100%\nTherefore, the probability that it is not windy is 100-12 = 88%
706951
I lost my grandmother at the airport. Her flight came in and I met her coming off the terminal but I turned around for a minute to look at something and she wandered off. It took me two hours to find her in some obscure store several terminals over. I could have prevented that by holding her hand all the way to baggage claim.
257322
It's very simple. Every taster has to choose one out of 5 cups, so on every round the probability of a single taster choosing A is 1/5.\nSince the event is repeated 4 times independently, the overal probability will be 1/5 of 1/5 of 1/5 of 1/5.\nThat is, (1/5)^4 = 1/625.
796216
First, choose lucky numbers.\nSecond, buy a ticket.\n\nIt really isn't complicated.
881071
the answer is 0.2875\nEXPLANATION:\n\nLow risk(LR) = 0.1,\npercentage Number of low risk Customers(PNLR) = 25% ( 0.25)\n\nHigh risk(HRLN) = 0.35,\npercentage Number of high risk customer = 75% ( 0.75)\n\nMr. Z can either be low risk or High risk because he was selected at random.\n\nProb. of Mr Z having accident =( LR x PNLR) or (HR x PNHR)\n                                                =(0.1 x 0.25) + ( 0.35 x 0.75)\n                                                = 0.2875
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!)
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.
75688
A confirmed flight ticket.......