4 quarters, 7 nickels, and 2 pennies in a wallet. One by one a coin is removed. This question came up in my textbook, and I simply couldn't get my head around it. The question is as follows:
What is the probability that the fourth coin that is taken out is a nickel and the seventh is a quarter?
I tried listing cases, but I soon found out that the process was far too tedious. Does anyone have any ideas as to how to solve this problem?
Another question that came from this is the following:
What is the probability that the last coin taken out is a penny?
I knew that there was some form of symmetry going on, and I could see why the solution would be $\frac{2}{13}$. However, I couldn't fully find a proof or some suitable reasoning behind this solution. Any help would be appreciated.
 A: the probability that a coin is drawn at the first, second, third....last draw is always the same, it cannot change.
Same reasoning for the rest of the questions.

You can verify my statement in an easy way:

*

*Penny drawn at first draw:

$$\frac{2}{13}$$

*

*Penny drawn at the second draw:

$$\frac{11}{13}\times \frac{2}{12}+\frac{2}{13}\times\frac{1}{12}=\frac{24}{13\times 12}=\frac{2}{13}$$
and so on...
A: In absence of any new information, the probability would remain the same as it was at the beginning.
One way to look at it is as below.
To start with, you have $4$ quarters, $7$ nickels, $2$ pennies (ratio of $4:7:2$).
How many quarters, nickels and pennies after first coin is taken out? You do not know but here are the expected number of coins based on their probability of being taken out.
Expected Number of Quarters $= 4 - \frac{4}{13} = \frac{48}{13}$
Expected Number of Nickels $= 7 - \frac{7}{13} = \frac{84}{13}$
Expected Number of Pennies $= 2 - \frac{2}{13} = \frac{24}{13}$
Their sum $= 12$ (coins left)
Their ratio based on expected numbers after one coin is taken out $= 4:7:2$ which is same as how you started. You can take these expected numbers and now try the second coin out but this ratio would not change.
A: None of the other answers attempted the first question and all only focused on the second question.
To reiterate, yes, the probability of whichever coin being pulled out in whichever position is going to be the same as the probability of that type of coin being pulled out in the first position.  See If you draw two cards what is the probability that the second card is a queen?
As for the first part of the question, what is the probability the fourth is a nickel and the seventh is a quarter, we use the same observation as well as the multiplication principle of probability noting that these events are not independent.
The probability the fourth is a nickel is $\frac{7}{13}$.  Given that the fourth is a nickel, the seventh is a quarter with probability $\frac{4}{12}$.  Multiplying these gives the probability of both happening together as being:
$$\frac{7}{13}\times\frac{4}{12}=\frac{7}{52}$$
