Answer to Question #124341 in Statistics and Probability for Yasini

Question

Suppose all licence plates in a certain region have three letters and three digits. If a plate is chosen at random. What is the probability that all three letters and all the numbers on the plate Will be different?

Answer 1

Solution.

Suppose there is $n$ letters and $k$ digits ($n \geq 3, k \geq 3)$, that can be on licence plate. And there are no any special rules, such as "digits 000 is forbidden".

Amount of all possible plates is $n^3\cdot k^3.$

Amount of plates with unique letters and digits is $n(n-1)(n-2)\cdot k(k-1)(k-2).$

So the probability is $p = \dfrac{n(n-1)(n-2)\cdot k(k-1)(k-2)}{n^3\cdot k^3}=\dfrac{(n-1)(n-2)\cdot(k-1)(k-2)}{n^2\cdot k^2} \\[0.1cm]$

If $n = 26$ and $k = 10$, then $p=\dfrac{25\cdot24\cdot9\cdot8}{26^2\cdot10^2}\approx 0.639.$

Answer.

0.639