I was solving the following problem:
There are three plain and four glazed donuts in a box; if two donuts are randomly eaten, what is the probability of randomly selecting a glazed donut if you know at least one of the eaten donuts was glazed?
I assumed that the first donut eaten was glazed, leaving 3 glazed and 3 plain donuts for the second choice, implying that there are two equally likely configurations that the third choice would be made from:
- two glazed and three plain (two glazed were selected)
- three glazed and two plain (one glazed and one plain were selected)
Giving a final probability of drawing a glazed donuts as $\left(\frac{1}{2}\right)\left(\frac{2}{5}\right)+\left(\frac{1}{2}\right)\left(\frac{3}{5}\right) = \frac{1}{2}$
However, when reviewing the solution, it was calculated, using conditional probability as follows:
$\begin{align}P(A \mid B)&=\frac{P(A \cap B)}{P(B)} \\ P(B)&=1-P(\bar{B}) =1-P(\{\text{plain},\text{plain}\}) \\ &=1-\frac{3}{7} \cdot \frac{2}{6}=\frac{36}{42} \\ P(A \cap B)&=P(\{\text{glazed}, \text{glazed}, \text{glazed}\} \cup\{\text{glazed}, \text{plain}, \text{glazed}\} \cup\{\text{plain}, \text{glazed}, \text{glazed}\})\\ &=\frac{4\cdot 3 \cdot 2}{7 \cdot 6 \cdot 5}+\frac{4 \cdot 3 \cdot 3}{7 \cdot 6 \cdot 5}+\frac{3 \cdot 4 \cdot 3}{7 \cdot 6 \cdot 5} = \frac{16}{35}\\ P(A \mid B)&=\frac{\frac{16}{35}}{\frac{36}{42}}=\boxed{\frac{8}{15}}\end{align}$
Where did I go wrong?
Selecting two glazed donuts or selecting one glazed and one plain donut is not equally likely:
There are $\binom{4}{2} = 6$ ways to select two glazed donuts while there are $4\cdot 3=12$ ways to select one glazed and one plain donut.
Hence, your calculation should be fixed as: $$ \frac{6}{18}\cdot \frac 2 5 + \frac{12}{18} \cdot \frac 3 5 = \frac{8}{15}. $$