So, with the remaining problems 2, 3, and 4, how do I show the probability by repeating digits? 2) 2773? 3) 8338? 4) 6444?
When Miguel starts his new job, what is the
1) 3278?
We know there are 4 different digits from 3278, and we know the total amount of digits is 0-9 which is a total of 10. So, working out problem one is shown the following below:
4!/10^4 = (4∙3∙2∙1)/10000 = 24/10000 = 3/1250 = 0.0024
So, with the remaining problems 2, 3, and 4, how do I show the probability by repeating digits?
2) 2773?
3) 8338?
4) 6444?
There are total 10 digit are possible(0-10) for each of last four digits.
Here we use Total Arrangement and Counting .
Probability = (No of Favorable Outcomes) / Total outcomes
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