It 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.
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.
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
3 times 40 = 120
14400.
Three full times. 120*3=360 390/120=3.25
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
10 times 12 equals 120.
120 times 8 is 960
5 times 24 = 120.
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.
12 x 10 = 120