Question

Illinois license plates used to consist of either three letters followed by three digits or two letters followed by four digits. Assume that all the different possible license plates are equally likely. (a) What is the probability that a randomly chosen plate contains the number 8888? (Round your answer to six decimal places.) (b) What is the probability that a randomly chosen plate contains the substring BE? (For example, BE4321 or PBE786 are two ways BE might appear. Round your answer to six decimal places.)

Answer 1

(a) 0.000028

(b) 0.002548

Explanation:

First, we find the total number of plates possible by adding the numbers of the two types of plates.

1) 3 letters followed by 3 digits

number of plates of type 1) = 26^3 * 10^3 = 17,576,000

2) 2 letters followed by 4 digits

number of plates of type 2) = 26^2 * 10^4 = 6,760,000

Total number of plates = 17,576,000 + 6,760,000 = 24,336,000

(a) Only type 2) plates can have the number 8888.

With a fixed number part of 8888, there are 2 letter positions that cna have 26 different letters each.

Number of type 1) plates with the number part 8888

26^2 = 676

P(8888 in plate) = 676/24,336,000 = 0.000028

(b) The substring BE can appear in type 1) plates in 2 ways:

xBE or BEx

Since x above can be one of the 26 letters, and since the string of letters is followed by 3 digits, there are 26 * 1000, or 26,000 possible plates for xBE and another 26,000 possible plates for BEx.

The substring BE can appear in type 2) plates in only 1 way, BExxxx, where xxxx is 10,000 combinations of digits, so there are 10,000 possible type 2) plates with the substring BE.

Total number of type 1) and type 2) plates with substring BE is

26,000 + 26,000 + 10,000 = 62,000

p(substring BE) = 62,000/24,336,000 = 0.002548