There are two things we need to recall in order to solve this problem.
1. The formula to find the probability:
[tex]P(E)= \frac{n(E)}{n(S)} [/tex]
where [tex]n(E)}[/tex] is the number of favorable outcomes of event E
and [tex]n(S)}[/tex] is the number of elements in the sample space
2. The Fundamental Principle of Counting (FPC). The FPC states that if one event can be done in [tex]m[/tex] ways and another event can be done in [tex]n[/tex] ways, the number of ways that the two events can happen is [tex]mn[/tex] ways. For example, if you have 2 different shirts and 5 different pair of pants, the number of possible combinations of a shirt and a pair pants that you can make is [tex](2)(5)=10[/tex] ways.
Now, the format of the plate number as described above can be illustrated this way:
________ ________ ________ ________ ________
(letter) (letter) (number) (number) (number)
(a) Let's find the sample space.
Since there are 26 letters in the alphabet and 10 digits to choose from, there are 26 choices of letters to put in the first blank, another 26 choices of letters to put on the second (the problem suggests that letters can be repeated), 10 digits on the third, 10 digits on the fourth, and 10 digits on the last blank (again, digits can also be repeated).
Using FPC, this can be translated to 26×26×10×10×10=676,000 ways, which is our sample space.
(b) Now, we find the number of ways that the license plate contains double letter and an even number.
For the letters, we can merge those two blanks as one because they have to be the same letter for both blanks and there are 26 ways that we can do that, i.e, AA, BB, CC, ..., YY, ZZ.
For the numbers, we remember two things: it has to be even and the digits can be repeated. So, we have 10 choices of digits on the 3rd blank, another 10 choices on the 4th blank, but only 5 choices of digits on the 5th blank (which are 0, 2, 4, 6, and 8) for the number to be even. (NOTE: Even 000 is an even number!)
Again, using FPC, the number of ways that the plate number can be done is 26×10×10×5=13,000 ways.
Therefore, the probability that the license plate can contain a double letter and an even number is
[tex]P(E)= \frac{13,000}{676,000}[/tex] or [tex]\frac{1}{52} [/tex].
