How to I solve this math question?
Number of license plate = 10^6
but I'm confused with the number of possible cases
You can have one '7' at 6 different places Then two '7' in 15 different cases C(6,2) Then three '7' in XX different case...
Same logic with '8' And a different logic with both '7' and '8'
But I have a hard time to figure out how to combine those logics to get the total number of possible cases.
Any help?
The first digit has a $\frac{8}{10}$ chance of not being a $7$ or $8$. The same goes for for the second, etc.
The chance that a number doesn't have any $7$ or $8$s at all is the combined chance that all digits are not a $7$ or $8$, this gives $\left(\frac{8}{10}\right)^6$.
Now the chance that a license plate does contain a $7$ or $8$ is simply $1 - \left(\frac{8}{10}\right)^6 = 73.7856\%$.