# Being the $n^{th}$ person to draw a coin from a bag of different coins

There is a bag that contains 20 silver coins and 10 gold coins. You are the sixth person in line to randomly draw and keep a coin from the bag.

a) What is the probability that you draw a gold coin?

b) If you draw a gold coin, what is the probability that the five people ahead of you all drew silver coins?

How do I account for the reduction in coins from each draw when it may be a gold or silver coin. Normally I could just reduce the probability of a collections of coins by reducing one, but since I am the sixth person I line how do I know what the number of gold and silver coins will be?

The probability the first person draws a gold is (1/3) and that they draw a silver is (2/3), but how does this influence the draw of the second person. Depending on what the first person drew (which is unknown) the second person will have different probabilities for gold and silver coins, but how do I account for that and continue that chain to the sixth person (me)?

For (a), it's an irrelevant distraction to think about the sequence of draws that come before you. Your coin has a $10$ in $30$ chance to be gold. If you need convincing, work it out for the second person instead of the sixth. For the second person, the chance of drawing gold is $$\frac{20}{30}\cdot\frac{10}{29}+\frac{10}{30}\cdot\frac{9}{29}=\frac{10}{30}\left(\frac{20}{29}+\frac{9}{29}\right)=\frac{10}{30}$$ It's not clear if you are asking for help with (b). But you can use $$\DeclareMathOperator{\P}{P}\P(\text{first five silver}\mid\text{sixth gold})=\frac{\P(\text{first five silver and sixth gold})}{\P(\text{sixth gold})}$$ where the denominator is your answer to (a) and the numerator is something you can calculate by multiplying a sequence of fractions.