What are the odds of drawing an even number less than 7 and an odd number greater than 10?
97799
i don't believe the person above me got the second part right.  he is right about the beginning - the first probability is 2/5:\n\npossible numbers to draw: 2 4 6 11 13 15\n\nso you can choose any of these six out of the fifteen given (6/15 = 2/5).  the next number you draw must be also a "winning" number.  BUT, you have to have an even number less than 7 AND an odd number greater than 10.  this gives us two scenarios:\n\nIF the first number picked was even and <7, then the possible numbers you can pick from are ONLY: 11 13 15, since picking another even number <7 will not meet your requirements.  this gives us a probablity of 3/14, since there are 3 "winning" numbers out of the 14 numbers left.\n\n\nIF the first number picked was odd and >10, then the possible numbers you can pick from are ONLY: 2 4 6, since picking another odd number >10 will not meet your requirements.  this gives us a probablity of 3/14, since there are 3 "winning" numbers out of the 14 numbers left.\n\nSO, finally, suppose you go with the first scenario.  the probability would be: (6/15)*(3/14) = 3/35  going with the second scenario will give you the same thing, 3/35.  thus adding these two, since you can have either scenario occur and still meet your requirements, gives you: 3/35+3/35 = 6/35.\n\nAnd that would be the odds: 6/35.
600596
There are 6 possible outcomes for the roll. There is only one outcome when the number is 4. Therfore the chances of getting a 4 are 1/6.
130268
To this day, it is not known if any odd perfect numbers exist, although numbers up to (10 to the 300th power) have been checked without success, making the existence of odd perfect numbers appear unlikely . There is a project underway at http://www.oddperfect.org/ seeking to extend the limit beyond (10 to the 300th power)\n\nCheck this link for more details:
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!)
1082362
Sometimes - rarely - there are.  I have very occasionally come to the end of the puzzle with two numbers that could go anywhere in two squares, ie, 2 and 6 might be interchangeable.  Then I think I'm wrong, and will do it all again to be faced with the same outcome!
360679
1, 2, 3, 4 are not greater than 4.  There are 6 numbers on a die.  4/6 or 66.6% chance that the number is not greater than 4.
279409
Find every number between 101 and 999 that is divisible by 7 and then add them together.  Hmmm......like I really am going to do that.
745507
diidy, how could the odds get worse when you buy more tickets?\n\nI think there are actually two answers, depending on whether each ticket is independent of the others or not.\n\nIf each ticket is independent, and there's no guarantee as to how many tickets there are or how many winners there will be, then think of it this way.  The odds of any ticket NOT winning are 299:300.  Therefore the odds of NO ticket winning are 299^n:300^n, where n is the number of tickets.  The odds of at least one ticket winning are 1 - the odds of no ticket winning, or 1 - 299^n:300^n, or (300^n - 299^n):300^n.\n\nIf there are 300 tickets of which 1 is a winner, then the odds of winning are n:300.
510570
Because the exponet of both is an even number which is 2\n\nIf you divide by 8 you might get a .5 result.
496127
For the number to be odd, the second digit must be 1, 3, or 5. Therefore, there are three choices for the second digit. The first digit can be any of the six digits, but it cannot be the digit used for the second digit. Therefore, there are 6 - 1 = 5 choices for the first digit.\n\nSo, 5 * 3 = 15.\n\nTo verify, the numbers are: E = 13, 15, 21, 23, 25, 31, 35, 41, 43, 45, 51, 53, 61, 63, 65. So, N(E) = 15 and the solution is correct.
239442
You're picking the wrong numbers. listen carefully, i'll tell you the winning numbers for last week 4-23-26-31-40-51.\nNo just joking, its a good idea to have three of the numbers to be even numbers and three to be odd numbers, these hit almost all the time, other than that it's all luck.
523765
1.  I'm assuming you mean "exactly one even card."\nP(1st card is even and 2nd card is odd and die is even) +\nP(1st card is odd and 2nd card is even and die is even)\n= (20/40)(20/39)(1/2) + (20/40)(20/39)(1/2) = 0.256\n\n2.  P(1st card is even and 2nd card is even and die is even)\n=  (20/40)(19/39)(1/2) = 0.123\n\n3.  Again, I'll assume you mean "exactly one card of 3 is drawn."\nP(1st card is 3 and 2nd card is not 3 and die is less than 3) + P(1st card is not 3 and 2nd card is 3 and die is less than 3)\n= (4/40)(36/39)(2/6) + (36/40)(4/39)(2/6) = 0.062\n\n4.  P(115 or 124 or 133 or \n142 or 151 or 214 or \n223 or 232 or 241 or \n313 or 322 or 331 or \n412 or 421 or 511) = \n(4/40)(3/39)(1/6) + (4/40)(4/39)(1/6) + (4/40)(4/39)(1/6) + \n(4/40)(4/39)(1/6) + (4/40)(4/39)(1/6) + (4/40)(4/39)(1/6) + \n(4/40)(3/39)(1/6) + (4/40)(4/39)(1/6) + (4/40)(4/39)(1/6) + \n(4/40)(4/39)(1/6) + (4/40)(4/39)(1/6) + (4/40)(3/39)(1/6) + \n(4/40)(4/39)(1/6) + (4/40)(4/39)(1/6) + (4/40)(4/39)(1/6)\n= 0.024\n\n5. P(111 or 112 or 121 or 211) =\n(4/40)(3/39)(1/6) + (4/40)(3/39)(1/6) + (4/40)(4/39)(1/6) + (4/40)(4/39)(1/6)\n= 0.0060
1179837
a) Try counting the possibilities.\n     There is only one way to roll a 12...(6, 6)\n     and there are 36 different possible rolls.\n     so the chances are 1/36\n\nb) There are 2 possibilities for this one (3, 6) and (6, 3)\n     so it's 2/36 or 1/18\n\nThe difference between the book the you is notation.\n1:35 means that for every one success there are 35 failures.\n1/36 means that there is one success for every 36 tries.
380874
Or 7:3 if the number of coins
90715
neither, the terms even and odd refer to integer numbers, not decimal or fractional ones.\n\nhope this helps :-)
861916
I can't see the rest of your question, but even numbers can be multiples of three. Like 6 (2x3). They are not mutually exclusive.
398144
Just multiply the numbers.  positives will get a positive answer.  Odd number of negatives will get a negative answer.  Even number of negatives will get a positive answer.