Using a probability tree to find the probability In a certain college, 55% of the students are women. Suppose we take a sample of two students. Use a probability tree to find the probability
a) that both chosen students are women.
b) that at least one of the two students is a woman. 
 A: Imagine that the drawing is done one at a time. So draw a tree. From the root come two branches. Label the node at the end of the left branch $W_1$, meaning the first person selected was a woman, and label the node at the end of the other branch $M_1$.
Write the relevant probabilities next to the line segment joining the root to each branch. The relevant probabilities are $0.55$ and $0.55$.
From the node labelled $W_1$, draw two branches, with the nodes at the end labelled $W_2$ and $M_2$, meaning that the second person selected is a woman (man). Do the same with the node labelled $M_1$.
Now we need to worry a bit about the probabilities to write next to the branches just drawn. For example, consider the branch joining $W_1$ to $W_2$. If the drawing is done with replacement, the answer is clear, it is $0.55$. 
If the sampling is done without replacement, things get more complicated. If the first person drawn is a woman, and the college is quite small, that one woman gone affects the probability that the next person drawn is a woman. However, if the college is large, the probability that the second person drawn is a woman given the first person was a woman is virtually equal to $0.55$. 
You are almost certainly expected to write $0.55$ next to the branch that joins $W_1$ and $W_2$. Then the branch that goes from $W_1$ to $M_2$ is $0.45$. 
Now to find the probability of two women, we follow the path that goes to $W_1$, then to $W_2$, and multiply the probabilities, getting $(0.55)^2$.
For at least one woman, we look at all paths that contain at least one woman, compute the probabilities of each, and add up. The path $W_1$ then $W_2$ has probability $(0.55)(0.55)$. The path $W_1$ then $M_2$ has probability $(0.55)(0.45)$. The path $M_1$ then $W_2$ has probability $(0.45)(0.55)$. Add.
It is much easier to realize that we are interested in all paths except the one that goes $M_1$ then $M_2$. That path has probability $(0.45)(0.45)$, so the probability of at least one woman is $1-(0.45)(0.45)$.  
