# Coin Draw Probability and Expected Value

Consider a pouch that contains seven gold coins and three silver coins. An experiment consists of drawing coins from a pouch until a silver coin is drawn. No coins are replaced. The outcome of the experiment is the number of coins drawn. For example, if the first coin was silver, then the outcome is 1 coin drawn.

I understand the possible outcomes of this scenario, but I can't figure out how to calculate the probability for each outcome. The possible outcomes range from 1 coin drawn to 8 coins drawn (all 7 gold are drawn before a silver one). I suppose the probability for 1 coin being drawn is 3/10, but I am not sure about the probabilities for the other outcomes.

For instance, would the probability of drawing 3 coins be (3/10)*(3/9)*(3/8)? If I calculate the probabilities that way, I get an expected value less than 1, which is impossible.

Let $X$ be the number of coins drawn until a silver coin encountered:

• $P(X=1)=\dfrac{3}{10}$

• $P(X=2)=\dfrac{7}{10}\cdot\dfrac{3}{9}$

• $P(X=3)=\dfrac{7}{10}\cdot\dfrac{6}{9}\cdot\dfrac{3}{8}$

• $P(X=4)=\dfrac{7}{10}\cdot\dfrac{6}{9}\cdot\dfrac{5}{8}\cdot\dfrac{3}{7}$

• $\dots$

• $P(X=N)=\dfrac{7!}{(8-N)!}\cdot\dfrac{(10-N)!}{10!}\cdot3$