There are 10,000 cars in a city, with license numbers from 00001 to 10

The pool of cars can be separated into those with at least one eight and those with no eights.

So, the number of cars with at least one eight can be determined by subtracting from the total number of cars the number with no eights.

The leading digit can be 0 or 1. The other digits can be 0-7 and 9, with the exception noted below.

So, the number of cars without 8's is:

9^4, which includes a nonexistent 00000 plate and excludes the 10000 plate, thus netting out.

So, the probability of not meeting a car with an eight is:

1-(9/10)^4

Two of the answer choices are very close to each other, suggesting that a lengthy multiplication is required.


The test makers are actually testing your ability to shortcut this process.

.9 to a power alternates between 9 and 1 as the furthest digit behind the decimal. So, with four as the power, that digit will be a 1.

Subtracted from 1 as above to determine the probability will leave a 9 as the digit rightmost.

That narrows the answer choices to A and D.

Answer A implies 9 raised to a power of 2 to be subtracted from 1, obviously incorrect, so the correct choice must be

D

Posted from my mobile device

Hi, thanks for the answer. I was trying to understand it from the Permutations & Combinations perspective. Could you please provide an answer using that?

There are 10,000 cars in a city, with license numbers from 00001 to 10,000.

When you meet a car, what is the probability there is a digit "8" in its license number ?

Total number of license numbers = 10000

Number of cases without digit 8 in its license number = 9^4 + 1 = 6562

Number of cases with digit 8 in its license number = 10000 - 6562 = 3439

The probability there is a digit "8" in its license number = 2439/10000 = .3439

IMO D