I'm trying to figure out what the odds of 1 in 8 will be as a percentage.  Can someone figure this out?
130215
one divided by 8 multiply 100
285500
Your logic is flawed. Say you always pick card #2. The first time, you'll have 14.2857...% (that's the same as saying one seventh) of the chances of winning the car.\n\nThe second time, with a new set of cards, your chances that the card #2 is the winner are again 14.2857...%, and so will every new time, because your chances are always 1 out of 7. You are picking one card out of seven cards that you are given.\n\nYour reasoning would be right if you could be assured that the car card will always be a different one (that for example, it will be in the first case #4, then #6, and so on). In that case, after the 7th. guess (if not earlier), you'd get the car. But the way you explained the game, nothing prevents the car from being always for example at card #5. If that is the case, always trying card #2 will make you fail every time.
83711
Id really like to know... why? but here goes\n7/8=.875 & that is 87.5%. 875%=8.75 and 875/100 which is 70/8
174209
If that one choice, 81/4 means 8.25, then that is the answer
208010
You need to know the total number of boys.\nThen take 4 divided by the total number of boys.\nThis will give you the probability.\nMultiply by 100 to get the percent.
75958
1/7 = 0.1428 = 14.28%\n(1 in seven has to be true)
388050
Probability(1st person picks any bill)=8/8\n\nProbability(2nd person picks the same bill)=1/8\n\nProbability(3rd person picks the same bill)=1/8\n\nThe events are independant as the outcome of one event will not affect the outcome of another. Thus allowing multiplication of probabilities.\n\nProb(All 3 draw same bill) = 8/8*1/8*1/8\n\n=1/64
50687
It's the same idea as what you did in the first one.\n\nThe 8 stays the same, and you reduce 8/12 to 2/3 so it is\n\n8 2/3.
217822
Yes i can say there are so many events that you can believe 50/50 chances.\n\nAs you said 50/50 is calculating for 100. means you have 2 chances.\n\n1. Toss the coin: to get head or tail you have 50/50 chance\n2. If a women is giving birth: the baby may be BOY/GIRL 50/50 Chance\n\n3. for this question : someone give the answer: which is correct/not correct   50/50 chance.\n........... \nyou can find somany events which are having 50/50 chances
586535
The GOOD news is, any combination of numbers has the exact same chance of winning as any other combination.  1-2-3-4-5-6 is just as likely to win as 12-4-98-32-5-7.  So, you needn't strain yourself trying to figure out which numbers to use.\n\nThe BAD news, of course, is that your chances of winning are still only about one in 15 million.  But hey, drop a buck on it if you want.  The gods may be trying to give you some money, so go ahead and make it easy for 'em.
379121
1.Probability of getting a number from 1-4 on throwing the dice\n         = 4/6\n\n2.Probability of a correct throw = 1/4\n\n3.Total number of questions = 160\n\nYour Answer = (4/6 * 1/4) raised to the power 160
215997
Here's a step by step explanation of how the probability is assessed.  Hope this will make things clearer for you.\n\nhttp://www.mste.uiuc.edu/java/java/birthday/explanation2.html
519615
I believe the answer is 44.44%.\n2/3 is the prob for heads and 1/3 is the prob for tails.\n\nThere are only two possible out comes to get this...heads then a tails...or tails then heads.\n\n2/3*1/3 is the outcome where you get a head then a tail.\nPlus 1/3*2/3 where you get a tail then a head.\n\nSo basically the answer is 2/3*1/3 + 1/3*2/3 or (2/3*1/3)*2 = 4/9
247098
33% is equal to approximately 3 out of 10(3.3)\nTherefore, if six out of ten were to be selected, that would equal 66% of ethnics - more than twice the probability.\nIt follows then that in iself, the odds would be just over two-to-one against this happening.\nSo the probability of more than of no more than six would be two tone ON.\nDoes that make sense?
258538
would calculate your chance of me calculating a statistics question for you at 1234.367:1
78034
(7/6-1)*1/6*100%=2.8%
397734
First you can establish the average result of a single dice by adding the probability of a result, multiplied by the odds of it occurring.\n\n(1 x 1/6)+(2 x 1/6)+(3 x 1/6)+(4 x 1/6)+(5 x 1/6)+(6 x 1/6)\n=\n1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6\n=\n21/6\n=\n3.5 (I.e. the avergage result on a single dice is 3.5)\n\nThen you simply divide your required result (15) by the average of a single dice (3.5) and round up to a whole number.\n\n15 / 3.5 = 4.29\nThis rounds up to 5.\n\nTherefore five dice are required to get an expected result of 15 or greater.