Question

A license plate is $3$ 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)

$\frac{7}{52}$

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 $\left(10\right)\left(10\right)\left(1\right)={10}^2$
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, $P\left(LorD\right)=P\left(L\right)+P\left(D\right)-P\left(LandD\right)$
$P\left(LorD\right)=P\left(L\right)+P\left(D\right)-P\left(L\right)P\left(D\right)$

$=1/26+1/10-\left(1/26\right)\left(1/10\right)$

$=10/260+26/260-1/260$
$=35/360=7/52$