There are 26 choices for the first letter and 26 choices for the second letter. Once the letters are chosen, there are 10 choices for each of the four digits. Therefore, the total number of possible plates is $26\times 26\times 10\times 10\times 10\times 10 = 67,\!600,\!000$.
To count the number of plates that have the letters at the beginning and at the end, we fix the first and last letters and multiply by the number of choices for the four middle digits. There are 26 choices for the first letter and 26 choices for the second letter, and 10 choices for each of the four digits. Therefore, the number of plates with the desired property is $26\times 10\times 10\times 10\times 26 = 67,\!600,\!000$.
Thus, the probability that a randomly chosen car has the letters at the beginning and at the end is $\frac{26\times 10\times 10\times 10\times 26}{26\times 26\times 10\times 10\times 10\times 10} = \boxed{\frac{1}{30}}.$
If a plate of a car consists of two letters and four digits and one car is chosen at random, then find the probability that the car has the letters at the beginning and at the end.
A 1/30
B 3/30
C 3/6
D 10/11
Good luck!!! Set by 😍😎 Abraham
1 answer