Note that repetition is allowed. This means a license plate can have three of the same letter and three of the same number.
New License Plates: 140 608 000 Probability: P(ALG-2)= 1175 760
Practice makes perfect
When picking the letters and numbers for the license plates, repetition is allowed. This means a license plate can have three of the same letter and three of the same number.
Example: AAA- 111
With this information we can analyze the number of possibilities there are in the original and new type of license plate.
Original License Plate
When picking the letters there are 26 options per letter. Similarly, there are 10 options per numbers. With this information we can write the following decision chart.
1 26*2 26*3 26*4 10*5 10* 6 10 = 17 576 000
There are 17 576 000 types of combinations when putting together the original license plate.
New License Plate
The second license plate has an additional digit ahead of the letters. Since this is limited to numbers from 1 to 9, we have 9 options to choose from. With this information we can calculate the number of license plates that can be formed.
1 9*2 26*3 26*4 26*5 10*6 10* 7 10 = 158 184 000
New - Original
By subtracting the number of original license plates from the number of new license plates, we can determine how many more license plates there are of the new type.
158 184 000-17 576 000=140 608 000
Chance of Getting ALG-2
To determine the probability of getting a plate with ALG-2 on it, we must divide the number of favorable outcomes with the total number of outcomes.
P=Number of favorable outcomes/Number of possible outcomes
We will assume that ALG-2 must come in exactly this order and that we are ordering a new license plate with the extra digit ahead of the letters. This means we have a total of 158 184 000 available plates that we can get. Let's list a few acceptable license plates.
3ALG- 289
7ALG- 228
4ALG- 291
This must mean that for the first number all 9 options are available. However, for the three letters and first number, we only have one available option. The last two digits can be whatever, so we have 10 options available for both of these picks.
1 9*2 1*3 1*4 1*5 1*6 10* 7 10 = 900
When we know how many license plates contain ALG-2, we can determine the probability of getting such a license plate.
P(ALG-2)=900/158 184 000 ⇒ 1/175 760
