Wiki User
∙ 10y agoIt is 1 - pr(no 5 nor 6 in 120 rolls)
= 1 - (2/3)120
= 1 - 1.113*1057
which is so close to 1 as to make no difference.
It is 0.99...98887 which starts off with a string of 56 nines.
Wiki User
∙ 10y agoIf you rolled a die 120 times, the probability of getting a 6 is one in six. It does not matter how many times you roll the die - the probability is still one in six - except that the long term mean will approach the theoretical value of 0.166... as the number of trials increases.
99/512, or 19.34%. The nCr formula can be used in this case: 12!/((12-7!)*7!) ---> 95,040/120 ---> 792 792/(2^12) = 99/512
120 = 12000
3 times 40 = 120
If it's a fair cube, then the probability of a ' 1 ' (or any other number 2 thru 6) onany roll is 1/6 . The most probable result of 120 rolls is 20 1s, 20 2s, 20 3s, 20 4s,20 5s, and 20 6s.That's the most probable. This means that if you do 120 rolls an infinite numberof times, that's the set of results you'll get most often. But it does NOT tell youwhat you WILL get if you do 120 rolls one time.A related comment: If you roll the cube 119 times and a ' 1 ' has never come upyet ... and it's a fair and honest cube ... then the chance of rolling a ' 1 ' on the120th roll is . . . . (wait for it) . . . . 1/6 , just like it was on the first roll. The cubehas no memory, and there's no such thing as "it's due".
If you rolled a die 120 times, the probability of getting a 6 is one in six. It does not matter how many times you roll the die - the probability is still one in six - except that the long term mean will approach the theoretical value of 0.166... as the number of trials increases.
The probability is 120/7776 = 0.0154, approx.
Because we are only modeling one event, all six outcomes of the die are equally possible. The probability of rolling a four (or, for that matter, any number) is 1/6, or .166666 repeating. Now, since we are modeling 120 rolls, the theoretical number of outcomes of four (or, again, any number) is 1/6 * 120 = 20 outcomes. The second sentence of the problem is unnecessary.
The probability of an even must be a real number in the range [0, 1]. 120% is outside that range.
99/512, or 19.34%. The nCr formula can be used in this case: 12!/((12-7!)*7!) ---> 95,040/120 ---> 792 792/(2^12) = 99/512
15 trials: 3 times 40 trials: 8 times 75 trials: 15 times 120 trials: 24 times But don't bet on it.
120
120 = 12000
There are 5! (that is, 120) distinct ways to arrange five items. Only 1 of them will have the books in alphabetical order by title. So the probability that it happens by random is 1/120.
10 times 12 equals 120.
120 times 8 is 960
5 times 24 = 120.