a combination lock is open with the correct sequence of 3 numbers from 0-27 inclusive. a number can be used more than once. find the probability of randomly choosing a sequence that will open the lock on the first try.
anyone know how to do this?
2006-11-05 18:22:49 · 7 answers · asked by fdsfsjk k 3 in Science & Mathematics ➔ Mathematics
First, if you are just going to have 3 numbers from 0 to 27 inclusively, you have 28 numbers all in all and it is with repitition so you will have just raise the 28 to the 3rd power
so you will get 21,952
and then , if you will have a correct answer on the first time you will get
1/21952
2006-11-06 00:54:30 · answer #1 · answered by jdash01 3 · 0⤊ 0⤋
The person named tma is correct. There is only one lock code out of all possible 28*28*28 codes
2006-11-05 18:36:33 · answer #2 · answered by Anonymous · 0⤊ 0⤋
the 1st extensive type must be 0 to 9, so there are ten a danger numbers. the 2nd extensive type is between 13 and 19, so there are 7 opportunities. The third extensive type is 30-39, 10 a danger numbers. for this reason the possibility would be a million out of (10 * 7 * 10), or a million out of seven-hundred.
2016-12-28 14:10:01 · answer #3 · answered by Anonymous · 0⤊ 0⤋
1/28^3
2006-11-05 19:13:18 · answer #4 · answered by fii 3 · 0⤊ 0⤋
since there are 28 numbers (0-27), i would think that there are 21,952 (28*28*28) possible combinations
so youre chances would be 1/ 21,952
i think
2006-11-05 18:29:43 · answer #5 · answered by tma 6 · 0⤊ 0⤋
probabilty = 1/(28^3) = 1/(28*28*28)
2006-11-05 19:49:53 · answer #6 · answered by lalalala 2 · 0⤊ 0⤋
I read the question wrong and gave you the wrong answer. tma is right.
2006-11-05 18:33:46 · answer #7 · answered by Biznachos 4 · 0⤊ 0⤋