Answer:
![P=0.0081](https://img.qammunity.org/2020/formulas/mathematics/middle-school/vu22tlc1w6kqchsasccn5fgv3izykn1fl9.png)
Explanation:
We know that the probability of an event is:
![P (event) = (Number\ of\ favorable\ outcomes)/(Number\ of\ possible\ outcomes)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/nvtpvx02ec2a2dizs0oy1ym6v7a4c2vegq.png)
Note that between 0 and 9 there are 10 possible digits {0,1,2,3,4,5,6,7,8,9}
If each digit can be repeated and the combination has 4 digits then the number of possible combinations S is:
![S = 10 * 10 * 10 * 10 = 10 ^ 4 = 10,000](https://img.qammunity.org/2020/formulas/mathematics/middle-school/hp8cf4oofnb68px40eekcw1qchork0qdbc.png)
If all digits obtained must be greater than six, then the possible digits that can be obtained are three: {7,8,9}.
If the combination is 4 digits then the number of results that can be obtained are:
![S = 3 * 3 * 3 * 3 = 3 ^ 4 = 81](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ukjcn7xwdmmobclnv9yuob535nuotvr5aw.png)
So:
Number of favorable outcomes = 81
Number of possible outcomes = 10,000
Finally the probability is:
![P = (81)/(10,000)\\\\P=0.0081](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ff4acq9dq2licagpwlokgd7dzk4q5apyne.png)