5.3 Probability of an Intersection

To find the probability that events \(A\) and \(B\) occur, we have to use the multiplication rule (i.e., probability of the intersection) which is written as follows: \[Pr(A \cap B) = P(A) \cdot P(B)\] For the multiplication rule to hold, the two events must be independent! The multiplication rule for dependent events will be introduced in more detail later but can be written as follows: \[Pr(A \cap B) = Pr(A) \cdot Pr(B|A)\] where \(Pr(B|A)\) is the probability of \(A\) given that even \(B\) occurred.

Suppose you have 16 polo shirts in your closet with your company’s logo. Nine of them are green and seven are blue. In the morning, you get dressed in the dark and randomly grab a shirt two days in a row (without doing laundry). What is the probability that both shirts are blue? Define the events \(B_i\) and \(G_i\) as grabbing a blue and green shirt, respectively on day \(i\). In this case, the probabilities are as follows: \[Pr(B_1) = 7/16\\ Pr(B_2|B_1) = 6/15\] Thus, \[Pr(B_1 \cap B_2) = Pr(B_1) \cdot Pr(B_2|B_1) = 7/16 \cdot 6/15 = 0.175\] If you are interested in the probability to get a 6 on roll 1 (event \(A\)) and a 6 on roll 2 (event \(B\)). This is written as \(Pr(A) \cdot Pr(B) = 1/6 \cdot 1/6 = 1/36\). This can also be expressed as a joint probability. The joint probability calculates the likelihood of two or more events happening at the same time.

Let \(A=\{Hearts\}\) and \(B=\{Queen\}\). Then the joint probability is the likelihood of drawing the Queen of Hearts from a deck of cards. This joint probability can be calculated as follows: \[Pr(A) = \frac{1}{4}\\ Pr(B) = \frac{4}{52}\\ Pr(A \cap B) = Pr(A) \cdot Pr(B) = \frac{1}{52}\] Let event \(A\) be that the first child is a girl and let event \(B\) be the second child being a girl. What is the sample space for the gender of the two kids? Given the sample space of \(\Omega = \{FF,FM,MF,MM\}\), we have \(Pr(A)=0.5\), \(Pr(B) = 0.5\), \(Pr(A \cup B)=0.5+0.5-0.25\), and \(Pr(A \cap B)=0.25\).