Probabilty

Question

How many 4 digit numbers can be formed from the digits 3, 7, 2, 5, using each digit only once.

 

What is the probability that the number 3 will be first or third?

 

What is the probability that the digits 7, 3, 2 will be the last digit?

Answer 1

Let's work on this problem in stages:

 

How many 4 digit numbers using 3, 7, 2, 5 without repetition (key words there):

 

This is a multiplication principle problem. We take each possibility and multiply them together.

The first digit has 4 possibilities.

Since we have removed one possibility from the pool in the first digit, we now only have 3 possibilities for the second digit.

The third digit has 2 remaining possibilities.

Which means the last digit only has one possibility left.

4 * 3 * 2 * 1 = 24

 

For 3 to be in either first or third positions, that means that 3 can be in half of the possible places. That means that we can use a permutation with 4 P 2, which winds up being 4!/(4-2)! = 24/2 or 12.

 

For the last part, I am assuming that any of the three digits is in the last position. If I'm wrong in this assumption, please let me know. We can do this problem backwards. By using the 4th digit in its place, we can use 1 - P(5) to find the probability of the 3 digits. The probability of 5 occuring in the last position is 24/4 = 6. So we can use 1 - P(5) to be 24 - 6 = 18 for the P(7,3,2).