Permutations and Combinations → Applications to Probability

Question

A four-digit number (with no repetitions) is to be formed from the set of digits {0, 1, 2, 3, 4, 5, 6, 7}. Find the probability that the number:

a is even 

b is odd 

c is less than 4000 

d is less than 4000, given that it is greater than 3000.

Answer 1

Firstly, we need to figure out how many numbers we could make. To do this, think of our 4-digit number as having 4 slots, and we need to fill each slot with a number from the set of digits.

When I fill the first slot (it doesn't matter which slot out of the 4 I pick), I have 8 digits I can choose from (0 through 7). I choose one at random, and move onto the next slot. Now, I only have 7 digits to choose from (let's say I picked the number 3 for the first slot, I can no longer use it since we don't allow repetitions). After picking a number for slot 2, I now only have 6 digits to choose from for slot 3. Lastly, after picking the number for slot 3, I only have 5 digits to choose from for slot 4.

Thus, we have 8 choices, then 7, then 6, and then 5, so in total, we have 8*7*6*5=1680 possible numbers we could make. This might seem too low, since 4 digits allows you to go all the way up to 9999, but remember that we only have access to digits up to 7, and also can never reuse digits, which removes a lot of numbers we could use (i.e. everything from 7700-7799).

a/b: Any number that ends in 0,2,4, and 6 will be even, and any number ending in 1,3,5, and 7 will be odd. When picking digits to put into our slots in the 4 digit number, every number is equally likely to end up in every slot, i.e. it's just as likely I pick 7 to go into the 1000's slot vs. I pick 2 to go into the 1000's slot. Similarly, since the numbers 1,3,5, and 7 are just as likely to end up in the 1's slot as are 0,2,4, and 6, we know that it's equally likely for the number to be odd as it is to be even. Thus P(even) = P(odd) = .5

c: Very similar logic to parts a/b. Any number that starts with 4,5,6, or 7 will be greater than 4000, and it is equally likely we put one of those into the 1000's slot as it is to put 0,1,2, or 3 there. Thus P(<4000) = .5.

d: Similar logic again! The only way for our randomly created number to be less than 4000 but greater than 3000 is for the 1000's slot to be filled by 3, and not any of the other numbers between 0 and 8. I said earlier that every number is equally likely to end up in any slot, so when picking the number for the first slot, there's a 1 in 8 chance of any number between 0-7 to end up in that slot. Thus, P(<4000 and >3000) = P(3 goes in the 1000's slot) = 1/8.