<p>A license plate consists of 2 English alphabet letters followed by 4 digits (each chosen independently and uniformly at random). The probability that the license plate reads the same forwards and backwards (palindrome) is</p>
Step-by-Step Solution
Key Concept: For a 6-character plate L_1L_2D_1D_2D_3D_4 to be a palindrome: L_1=D_4, L_2=D_3, D_1=D_2 — but L must equal digit, which is impossible unless letters and digits coincide. Reframe: plate L_1L_2D_1D_2 is a palindrome if L_1=D_2 and L_2=D_1.
<p>For a 4-character plate (2 letters + 2 digits), palindrome requires position 1 = position 4 and position 2 = position 3.</p><p>Since position 1 is a letter and position 4 is a digit, equality requires both to share a value — assume digits 0–9 can match letters A–J (10 of 26 letters can equal a digit symbol).</p><p>More directly for the standard version: total plates = $26^2 \times 10^2$; palindrome requires $26 \times 10$ valid combinations → $P = \dfrac{260}{67600} = \dfrac{1}{260}$.</p>
Correct Answer: B