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Q. 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

Probability - Part 2

Solution:

Let A : event that the place has a three letter palindrome $B$ : event that the place has a three digit palindrome
$P ( A )=\frac{26^2}{26^3}=\frac{1}{26} \quad\left( L _1 L _2 L _1\right) \text {; ||| ly } P ( B )=\frac{10^2}{10^3}=\frac{1}{10}$
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hence, $P ( A$ or $B )= P ( A )+ P ( B )- P ( A \cap B )=\frac{1}{26}+\frac{1}{10}-\frac{1}{26 \cdot 10}=\frac{10+26-1}{260}=\frac{7}{52}$