A palindromic number is a number that remains the same when its digits

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4-digit palindromic numbers have the form \(‘xyyx’\), where
\(x\) is one of values \(1,2,…,9\) and \(y\) is one of values \(0,1,2,…,9.\)

So, there are \(9 * 10 = 90\) four-digit palindromic numbers.
The total number of 4-digit numbers between \(1000\) and \(9999\), inclusive, is \(9000 ( = 9999 – 1000 + 1 )\).

Therefore, the probability that the selected 4-digit is palindromic is \(\frac{90}{9000} = \frac{1}{100}.\)

Therefore, the answer is E.
Answer: E

Let’s determine the number of the 4-digit palindromes. Notice that a 4-digit number is a palindrome if it’s one of the following two formats: XXXX and XYYX where X and Y represent a digit and X ≠ Y and X is nonzero.

Format 1: XXXX

We see that X can be any digit from 1 to 9, inclusive; thus, there are 9 such numbers.

Format 2: XYYX

We see that X can be any digit from 1 to 9, inclusive, and Y can be any digit from 0 to 9, inclusive (excluding digit X), so that there are 9 choices for X and 9 choices for Y; thus, the number of 4-digit number in this format is 9 x 9 = 81.

Thus, there are a total of 9 + 91 = 90 numbers in both formats. Since there are 9000 four-digit numbers (1000 to 9999 inclusive), the probability of picking a 4-digit palindrome randomly is 90/9000 = 1/100.

Answer: E

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