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
To find the probability that the car has the letters at the beginning and at the end of the plate, we need to determine the total number of possible car plates, as well as the number of car plates that have letters at the beginning and at the end.
Since the plate consists of two letters and four digits, there are 26 choices for each letter (from A to Z) and 10 choices for each digit (from 0 to 9).
The total number of possible car plates is therefore calculated as:
Total number of plates = (Number of choices for the first letter) * (Number of choices for the second letter) * (Number of choices for the first digit) * (Number of choices for the second digit) * (Number of choices for the third digit) * (Number of choices for the fourth digit)
= 26 * 26 * 10 * 10 * 10 * 10
= 676,000
Now, let's determine the number of car plates that have letters at the beginning and at the end. Since there are 26 choices for each of the two letters, the number of plates that meet this condition can be calculated as:
Number of plates with letters at the beginning and at the end = (Number of choices for the first letter) * (Number of choices for the second letter) * (Number of choices for the first digit) * (Number of choices for the second digit) * (Number of choices for the third digit) * (Number of choices for the fourth digit)
= 26 * 26 * 10 * 10 * 10 * 10
= 676,000
Therefore, the probability that the car has the letters at the beginning and at the end is:
Probability = (Number of plates with letters at the beginning and at the end) / (Total number of plates)
= 676,000 / 676,000
= 1
Thus, the correct answer is not provided among the options given.
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 $\frac{26\times 10\times 10\times 10\times 26}{26\times 26\times 10\times 10\times 10\times 10} = \boxed{\frac{1}{30}}.$