Question

A license plate is 3 capital letters (of English alphabets) followed by 3 digits. If all possible license plates are equally likely, the probability that a plate has either a letter palindrome or a digit palindrome (or both), is

• A. $\frac{7}{52}$
• B. $\frac{9}{65}$
• C. $\frac{8}{65}$
• D. none

Answer 1

Answer: (A)

Let A : event that the place has a three letter palindrome $B$ : event that the place has a three digit palindrome
$P(A)=\frac{26^2}{26^3}=\frac{1}{26}\left(L_1{L}_2{L}_1\right)\text{; ||| ly }P(B)=\frac{10^2}{10^3}=\frac{1}{10}$

hence, $P(A$ or $B)=P(A)+P(B)-P(A\cap B)=\frac{1}{26}+\frac{1}{10}-\frac{1}{26\cdot 10}=\frac{10+26-1}{260}=\frac{7}{52}$