Question

Many states use a sequence of three letters followed by a sequence of three digits as their standard license-plate pattern. Given that each three-letter three-digit arrangement is equally likely, what is the probability that a ran- domly chosen license plate contains at least one palindrome (a three-letter arrangement or a three-digit arrangement that reads the same left-to-right as it does right-to-left)?

Answer 1

Explanation:

There are 26 letters, so there are 26

3

possible 3-letter sequences.

the number of these which are palindromes can be computed as follows:

26 choices for the first letter

26 choices for the second letter

1 choice for the third letter ( it has to agree with the first)

number of ways =26

2

Probability of a letter palindrome =26

2

/26

3

=1/26

there are 10 digits, so there are 10

3

sequences of digits.

Reasoning as above, the number of digit-palindromes is (10)(10)(1)=10

2

so the probability of a digit-palindrome is 10

2

/10

3

=1/10

The events

D - digit palindrome

L - letter palindrome

are independent but not mutually exclusive.

So, P(LorD)=P(L)+P(D)−P(LandD)

P(LorD)=P(L)+P(D)−P(L)P(D)

=1/26+1/10−(1/26)(1/10)

=10/260+26/260−1/260

=35/360=7/52