Question

consider all the five digit numbers that can be created from the 0 -9 where the first and last digits must be odd and no digit can repeat. What is the probability of choosing random number that starts with 3 from this group ? Enter as a fraction or round the answer to 4 decimal places

Answer 1

The number will have 5 slots. The first digit can be any odd digit from 0 to 9, then the first digit is one of the numbers in the set {1,3,5,7,9}, this means that we have 5 options. Now the last digit can't be the same as the first one but it has to be odd as well. then we have 4 options for the last slot.

Now from the second, third and fourth slots we can chosse any digit from 0 to 9 as long as it is not repeated then we have 8, 7 and 6 options. Then the total amount of 5 digit numbers with this specifications is:

$5\cdot8\cdot7\cdot6\cdot4=6720$

Now from this total the amount that starts with 3 is:

$1\cdot8\cdot7\cdot6\cdot4=1344$

Therefore the probability is:

$\frac{1344}{6720}=\frac{1}{5}$

and this is equal to 0.2