Tough probability question?
967346
This probability problem can be used using Binomial Distribution.  \n\nGiven a number of 'x' success, the probability in a binomial experiment is:\n\nP(x) = [n choose x]*p^x*q^(n-x)\n\nwhere:\n\nn choose x is a combination problem: n!/[(n-x)!x!]\np = probability of success\nq = probability of failure\nn = total selection\nx = successes\n\nIn this problem, we are looking for at least one case has a success (which is one of the highest 9 numbers).  In essence, we are solving for:\n\nP(x=1) + P(x=2) + P(x=3) + P(x=4) + P(X=5).  \n\nHowever it would be easier to find the probability of the failures and subtract from the total probability because we only have to calculate 1 probability.  This would be:\n\n1 - P(x=0) is what we are looking for.  Now all we need is to calculate the probabilty.\n\nFor P(x=0), we have the following:\n\nn = 5\nx = 0\np = (9/26) = probability of success\nq = (17/26) = probability of failure\n\nPlugging into P(x), we get:\n\nP(x=0) = (5!/5!)(9/26)^0(17/26)^5\n\nSolving, you get P(x=0) = 0.12\n\nTherefore, \n\n1 - P(x=0) is ------>\n\n1- 0.12 = 0.88\n\nBelow is a website devoted to Binomial Distribution exercises\n---------\n\nHope this helps
405425
Probability of getting one head = 1/2\nProbability of getting two heads = (1/2) * 2 = 1/4\nYou have a 25% chance of getting a head two times...\nAnd 12.5% chance of getting a head 3 times...\nAnd 6.25% chance of getting a head 4 times...\n.\n.\n.\nAnd 0.09765625% chance of getting a head 10 times!!! :)
90357
F- even and odds.  There are 3 even sides and 3 odd sides. So the likelihood is the same for both.
770080
Let us call x the minutes hand on the clock.\nIf 0<x<=5, then it is train B. (5 minutes)\nIf 5<x<=12, then it is train A. (7 minutes)\nIf 12<x<= 24, then it is train A. (12 minutes)\nIf 24<x<= 25, then it is train B. (1 minute)\nIf 25<x<= 45, then it is train B. (20 minutes)\nIf 45<x<=48, then it is train A. (3 minutes)\nIf 48<x<=60, then it is train A. (12 minutes)\n\nWe add it up to get that for 34 of the minutes, he gets on A, and that for of the 26 minutes he gets on B. There are 60 minutes in an hour, so the probability is 34/60, which reduces to 17/30.
877981
For the first part, per hour the rate is 30.  so we have lambda = 30, so we know the mean and variance (look at the answer before mine).  To find the Probability the 240th person enters in the 8th hour, find the probability 240 or more people arrive in an 8 hour day, then times that by the probability less than 240 arrive in 7 hours.\n\nso (1-P(<240 in 8))*P(<240 in 7)\n\nRemember, The poisson process is memoryless.\n\nSteve
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.
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.
832564
The answer is one chance in 2^5  or 1 in 32.\n\nNotice that the chances are the same as flipping five heads in a row.
192695
You have the following lists:\nB B B\nB B G\nB G B\nG B B\nB G G\nG B G\nG G B\nG G G\n\nSo you have 8 variations and of those 3 are 2 boys 1 girl so the probability is 3/8.
1136735
P(Last is W) = P{(2nd is W and 1st is W) or (2nd is B and 1st is W)}\n= P(2nd is W and 1st is W) + P(2nd is B and 1st is W) (since these are mutually exclusive events)\n= P(2nd is W)P(1st is W|2nd is W) + P(2nd is B)P(1st is W|2nd is B) (both use the general multiplication rule)\n= (6/10)(7/11) + (4/10)(6/11)\n= (42+24)/110\n= 66/110\n= 0.6\n\nWhich is the probability of getting a white from either box, which I think makes sense, since both have the same proportion of whites.
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.
795924
Binomial distribution, (a) p=0.5, n=12; (b) p=0.5, n=13.  You can do it!
258367
.001% Very small probability.
714109
When there are 80 numbers with equal chance the probability is 1 / 80 for each number.\nThe chance of 20 out of 40 numbers is 1 out of \n40 x 39 x 38 x... x 22 x 21 = 40! / 20! = 3 x 10^29\nSince you have all the 80 numbers, the chance is 4 x 10^27.\nThe very, very, very most probability is that you waist your money again\n\nTh
850369
The answer to this question is (b) 2/7.\n\nWhen removing two dogs there are four possible scenarios:\n\n(1) Remove poodle, then remove terrier.\n(2) Remove poodle, then remove another poodle.\n(3) Remove terrier, then remove another terrier.\n(4) Remove terrier, then remove poodle.\n\nIn each case, take the probability of the first even happening and multiply it by the probability of the second even happening. For instance:\n\nProbability that a poodle is first selected = 5(# of poodles originally) / 7(total # of dogs).\nProbability that a terrier is then selected = 2(# of terriers left) / 6(total # of dogs after a poodle is removed).\nMultiply 5/7 and 2/6 to get 10/42.\n\nRepeat this with each of the remaining three cases to get\n(1) 10/42\n(2) 20/42\n(3) 2/42\n(4) 10/42\n\n(Note that these add up to 1, which means that we have all of the possible cases)\n\nNow, just add the probabilities of the two cases with a terrier being selected second to get:\n\n10/42(first case) + 2/42(third case) = 12/42\n\nReduce 12/42 down to 2/7 to get your answer.\n\nI hope this helps, good luck with future probability problems!
466671
Flipping a coin 4 times means that there can be 16 different posibilites of Heads and Tails occuring (2^4 = 16).\n\nThus by dividing 1 (max. probability) by 16 we get .0625. So each combonation has a probability of 6.25%. Now there is only one possibilty (TTTT) where there are no heads thus the possibility of no heads is simply .0625 or 6.25%\nWhere as one head has the following possibilites:\n\nTTTH\nTTHT\nTHTT\nHTTT\n\nThus multiplying .0625 by 4 we get .25 which is possibilty of only 1 head occuring in four flips.\n\nSo in conlusion:\n\nFor four flips of a coin:\n\nProbability of only one head occuring = .25 or 25%\nProbability of no head occuring       = .0625 or 6.25%
185212
Probability that the first coil is good = 88/100\nProbability that the second coil is good = 88/100 (irrespective of the first choice)\n\nProbability that both are good = (88/100)^2 = 77.44%