Suppose license plates have 6 entries. The first entries must be letters (A-Z),

Question

Suppose license plates have 6 entries. The first entries must be letters (A-Z), and the last three must be digits (0-9). Letters and digits can be used more than once.

a.) A license plate is selected randomly. What is the probability that the letter Z is not used?

b.) Jen was born in Februrary of 88. She chooses a license plate by randomly ordering the letters J, E, and N in the first 3 spots and randomly ordering the digits 2, 8, and 8 in the last 3 spots. What is the probability that Jen chooses a license plate with 2 adjacent 8's?

Explanation / Answer

a) There are 26 letters. Number of ways of putting in the first 3 entries is 26*26*26 (since repetition is allowed) = 26³
There are 10 digits, so number of ways of filling the last 3 entries is 10*10*10 = 10³

So, total number of possible license plates is the product of the above two = 260³

Now, if Z is not used, we have 25 letters to select from. So, the number of ways of putting in first 3 entries becomes 25³ The digits part remains unchanged as far as number of ways go.
So, total possible license plates without Z = 250³

Required probability = 250³/260³ = (25/26)³

b) The first 3 letters can be scrambled in 3! = 6 ways.
The last 3 digits can be arranged in 3!/2! = 3 ways (Make consideration for the identical 8)
Total number of possible license plates = 6*3 = 18

Now, the 8's are adjacent in only 2 of the 3 possible digit arrangements (882 and 288). Number of letter arrangements in unchaged (6).
So total plates with adjacent 8's = 6*2 = 12

Required probability = 12/18 = 2/3

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