A bag contains 3 red jewels and 7 black jewels. You randomly draw the jewels out one by one without replacement. What is the probability that the last red jewel was the 8th one withdrawn?

We have total of 10 balls or jewels, say balls. We can order them in $10!$ ways. 

If we want the last ball to be red, then we choose one red ball in ${3 \choose 1}=3$ ways, put it aside and list the remaining 9 balls in $9!$ ways, and put the chosen red ball at the end of the list. So the number of favorite possibilities is

\[9! \times 3.\]

Hence the probability that the last red ball was the 8th one withdrawn is 
\[\frac{9! \times 3}{10!}=\frac{3}{10}.\]
What I have compted is the probability that the last ball  (10th ) is red. 

The solution to your problem: 

First choose two black balls that will be the last two balls which should be black. This could be done in
\[{7 \choose 2}=21\]
ways. We can order these two balls in $2$ differnt ways to form the last 2 black balls in the list. 

Now choose the 8th ball that is supposed to be red, we can choose one red ball in ${3 \choose 1}=3$ ways. So we can form the last  three balls in the list in $3 \times 21 \times 2=126$ different ways. The remianing 7 balls can be listed in $7!$ ways to form the first 7 balls. So the probability is

\[\frac{7! \times 126}{10!}=\frac{126}{10 \times 9 \times 8}=0.175.\]