From the digits 0, 1, 2, … 9 randomly choose three digits. What is the probability that these three digits can be used to write:
a) at least one three-digit odd number
b) exactly one three-digit odd number
c) exactly two three-digit odd numbers
Here is my try:
a) $1 – \frac{\binom{3}{5}}{\binom{3}{10}} = \frac{11}{12}$
This doesn't lead me to a solution of b) but I tried to take one odd digit and any two of the remaining digits dividing by all 3 choices of digits $\frac{\binom{1}{5} * \binom{2}{9}}{\binom{3}{10}}$ to get probability of at least one odd and any two digits but it seems to be $\frac{3}{2}$. Help me understand what is wrong.
Update:
b) Now I realize that in order to get exactly one three-digit odd number i need to have 0, any odd and any even digit, so it will be $\frac{\binom{1}{5} * \binom{1}{4}}{\binom{3}{10}} = \frac{1}{6}$
c) I need 0, and two odd digits then probability will be $\frac{\binom{2}{5}}{\binom{3}{10}} = \frac{1}{12}$, it matches the answer but I still have my doubts
Best Answer
Question b)
There are four cases: CASE 1 you have no odd numbers (p = 1/12), CASE 2 you have one odd number (p=5/12), CASE 3 you have two odd numbers (p=5/12), CASE 4 you have three odd numbers (p=1/12).
The only way in which you can create exactly one three-digit odd number is in CASE 2 when one of the two even numbers is 0. The probability that one of the even numbers is 0, is $\frac{4}{2 \choose 5} = \frac{2}{5}$. So the probability that you can create a three-digit odd number equals $\frac{5}{12}.\frac{2}{5} = \frac{1}{6}$
Question c)
This can only happen in CASE 2 (p=5/12) where you have one odd number and additionally none of the other numbers are zero (p = 3/5), or in CASE 3 (p=5/12) where you additionally have one zero (p=1/5). this combined yields a probability of $\frac{5}{12}.\frac{3}{5} + \frac{5}{12}.\frac{1}{5} = \frac{1}{3}$