Section 4.5 Conditional Probability
When finding the probability of an event, sometimes you may need to consider past history and how it might affect things. Indeed, you might think that when the local station forecasts rain then the probability of it actually raining should be greater than if they forecast fair skies. At least that is the hope. :) In this section, you will develop a way to deal with the probability of some event that might change dependent upon the occurence or not of some other event.
Indeed, consider what happens when you keep on dealing a hand of five cards from a shuffled deck but without replacement. Notice how the probability of the same thing (such as P(getting a Heart on the next card)) oscillates based upon what cards came out of the deck on previous hands.
Now, consider the case when you put the cards back in, reshuffle, and then get 5 new cards...
Consider a box with three balls: one Red, one White, and one Blue. Using an equally likely 4.3.11 assumption, the probability of randomly pulling out a Red ball should be 1/3. That is P(Red) = 1/3.
However, suppose that for a first trial you pull out the White ball and set it aside. Attempting to pull out another ball leaves you with only two options and so the probability of randomly pulling out a Red ball is 1/2. Notice that the probability changed for the second trial dependent on the outcome of the first trial.
Example 4.5.2. Changing Sample Space - Cards.
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*}
Definition 4.5.3. Conditional Probability.
For sets A and B,
\begin{equation*}
P(B | A) = \frac{P(A \cap B)}{ P(A) },
\end{equation*}
provided \(P(A) \gt 0\text{.}\)
You can read \(P(B|A)\) as "the probability of B given A".
Theorem 4.5.4.
Conditional Probability 4.5.3 satisfies all of the requirements of regular probability 4.3.2.
Proof.
By definition, for any event probability must be nonnegative. Therefore
\begin{equation*}
P(A \cap B) \ge 0.
\end{equation*}
So,
\begin{equation*}
P(B | A) = \frac{\text{positive or zero}}{\text{positive}}\ge 0.
\end{equation*}
Further,
\begin{equation*}
P (S | A) = P(A \cap S)/P(A) = P(A)/P(A) = 1.
\end{equation*}
For the third part, we will only consider the case when there are two disjoint sets B and C. Then,
\begin{align*}
P(B \cup C | A) & = \frac{P(A \cap (B \cup C)}{P(A)} \\
& = \frac{P( (A \cap B) \cup (A \cap C) )}{P(A)}\\
& = \frac{P(A \cap B)}{P(A)} + \frac{P(A \cap C)}{P(A)}\\
& = P(B | A) + P(C | A).
\end{align*}
Theorem 4.5.5. Multiplication Rule.
For any sets A and B,
\begin{equation*}
P(A \cap B) = P(A) P(B | A) = P(B) P(A | B)
\end{equation*}
Proof.
If \(P(A)=0\) or \(P(B)=0\text{,}\) then the result is trivial. Otherwise, unravel the definition of conditional probability 4.5.3 by taking the denominator to the other side. Also note that you can write \(A \cap B = B \cap A\text{.}\)
Checkpoint 4.5.6. WebWork - Disjoint Pieces.
A bag contains \(4\) red marbles and \(5\) white marbles. Two marbles are drawn in succession without replacement. Find the probabilities of the following events:
1. The first marble drawn is red and the second is white.
Answer:
2. Both marbles drawn are red.
Answer:
See how you had to break the given question up into two disjoint pieces.
