Consider a deck of 52 standard playing cards and a success occurs when a Heart is selected from the deck. When extracting one card randomly, the probability of that card being a Heart is P(Heart) = 13/52.

Now, assume that one card has already been extracted and set aside. Next, prepare to extract another. If the first card drawn was a Heart, then there are only 12 Hearts left for the second draw. However, if the first card drawn was not a Heart, then there are 13 Hearts available for the second draw. To compute this probability correctly, one need to formulate the question so that subadditivity can be utilized.

Let \(H_1\) be the outcome Heart on 1st draw and \(H_2\) be the outcome Heart on 2nd draw. Then,

\begin{align*} P(\text{Heart on 2nd draw}) & = P( [ H_1 \cap H_2 ] \cup [ H_1^c \cap H_2 ] )\\ & = P( H_1 \cap H_2 ) + P( H_1^c \cap H_2 )\\ & = \frac{ | H_1 \cap H_2 |}{| P( \text{Number of ways to get two cards} | }\\ & + \frac{ | H_1^c \cap H_2 | }{ | \text{Number of ways to get two cards} | }\\ & = \frac{13}{52} \cdot \frac{12}{51} + \frac{39}{52} \cdot \frac{13}{51} = \frac{12}{4 \cdot 51} + \frac{3 \cdot 13}{4 \cdot 51} \end{align*}