You are given a random 4 digit code where all integers from 0 to 9 are valid options. O and 9 are included in this range. Additionally, digits may be repeated. With this information, solve the following three problems. What is the probability that the number in the 4ths integer's spot is a 6? What is the probability that you are given two 5s? What is the probability you are given three 5s?
Given,
- A 4 digit code is formed by using the integer from 0 to 9.
- Repetition of digits are allowed
The first place of the 4 digit code can be filled in 10 ways by using the digits 0 to 9.
Since the repetition of digits are allowed,
The first place of the 4 digit code can be filled in 10 ways by using the digits 0 to 9.
The second spot of the 4 digit code can be filled in 10 ways by using the digits 0 to 9.
The third spot of the 4 digit code can be filled in 10 ways by using the digits 0 to 9.
The fourth spot of the 4 digit code can be filled in 10 ways by using the digits 0 to 9.
By using the counting principle of permutation, the total number of 4 digit code can be formed is,
| 6 |
If the fourth spot is fixed by the integer 6, then the total number of 4 digit code can be formed is,
Therefore, the probability of that the number 6 in the fourth spot can be calculated as,
Hence, the probability of that the number 6 in the fourth spot is 0.1 .
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