# Conditional Probability

# Introduction

One of the key properties of coin flips is independence: if you flip a fair coin ten times and get ten \(T\)’s, this does not make it more likely that the next coin flip will be \(H\)’s. It still has exactly \(50\%\) chance of being \(H\)’s.

By contrast, suppose while dealing cards, the first ten cards are all red (hearts or diamonds). What is the chance that the next card is red? This is easy: we started with exactly \(26\) red cards and \(26\) black cards. But after dealing the first ten cards we know that the deck has \(16\) red cards and \(26\) black cards. So the chance that the next card is red is \(16/42\). So unlike the case of coin flips, now the chance of drawing a red card is no longer independent of the previous card that was dealt. This is the phenomenon we will explore in this note on conditional probability.

# Conditional Probability

Let’s consider an example with a smaller sample space. Suppose we toss \(m=3\) balls into \(n=3\) bins; this is a uniform sample space with \(3^3=27\) points. We already know that the probability the first bin is empty is \((1-1/3)^3=(2/3)^3=8/27\). What is the probability of this event *given that* the second bin is empty? Call these events \(A,B\) respectively. In the language of conditional probability we wish to compute the probability \({\mathbb{P}}[A\mid B]\), which we read to say probability of \(A\) given \(B\).

How should we compute \({\mathbb{P}}[A\mid B]\)? Well, since event \(B\) is guaranteed to happen, we need to look not at the whole sample space \(\Omega\), but at the smaller sample space consisting only of the sample points in \(B\). In terms of the picture below, we are no longer looking at the large oval, but only the oval labeled \(B\):

What should the probabilities of these sample points be? If they all simply inherit their probabilities from \(\Omega\), then the sum of these probabilities will be \(\sum_{\omega\in B}{\mathbb{P}}[\omega]={\mathbb{P}}[B]\), which in general is less than \(1\). So we need to *scale* the probability of each sample point by \(1/{\mathbb{P}}[B]\). I.e., for each sample point \(\omega\in B\), the new probability becomes \[{\mathbb{P}}[\omega\mid B]={{{\mathbb{P}}[\omega]}\over{{\mathbb{P}}[B]}}.\] Now it is clear how to compute \({\mathbb{P}}[A\mid B]\): namely, we just sum up these scaled probabilities over all sample points that lie in both \(A\) and \(B\): \[{\mathbb{P}}[A\mid B] = \sum_{\omega\in A\cap B} {\mathbb{P}}[\omega\mid B]
= \sum_{\omega\in A\cap B} {{{\mathbb{P}}[\omega]\over{{\mathbb{P}}[B]}}}
= {{{\mathbb{P}}[A\cap B]}\over{{\mathbb{P}}[B]}}.\]

**Definition 1** *(Conditional Probability)*

*For events \(A,B\) in the same probability space, such that \({\mathbb{P}}[B]>0\), the conditional probability of \(A\) given \(B\) is \[{\mathbb{P}}[A\mid B] = {{{\mathbb{P}}[A\cap B]}\over{{\mathbb{P}}[B]}}.\]*

Returning to our example, to compute \({\mathbb{P}}[A\mid B]\) we need to figure out \({\mathbb{P}}[A\cap B]\). But \(A\cap B\) is the event that both the first two bins are empty, i.e., all three balls fall in the third bin. So \({\mathbb{P}}[A\cap B]=1/27\) (why?). Therefore, \[{\mathbb{P}}[A\mid B] = {{{\mathbb{P}}[A\cap B]}\over{{\mathbb{P}}[B]}} = {{1/27}\over{8/27}} = {1\over 8}.\] Not surprisingly, \(1/8\) is quite a bit less than \(8/27\): knowing that bin \(2\) is empty makes it significantly less likely that bin \(1\) will be empty.

## Example: Card Dealing

Let’s apply the ideas discussed above to compute the probability that, when dealing \(2\) cards and the first card is known to be an ace, the second card is also an ace. Let \(B\) be the event that the first card is an ace, and let \(A\) be the event that the second card is an ace. Note that \({\mathbb{P}}[A] = {\mathbb{P}}[B] = 1/13\).

To compute \({\mathbb{P}}[A\mid B]\), we need to figure out \({\mathbb{P}}[A\cap B]\). This is the probability that both cards are aces. Note that there are \(52\cdot 51\) sample points in the sample space, since each sample point is a sequence of two cards. A sample point is in \(A\cap B\) if both cards are aces. This can happen in \(4\cdot 3 = 12\) ways.

Since each sample point is equally likely, \({\mathbb{P}}[A\cap B] = 12/(52 \cdot 51)\). The probability of event \(B\), drawing an ace in the first trial, is \(4/52\). Therefore, \[{\mathbb{P}}[A\mid B] = {{{\mathbb{P}}[A\cap B]}\over{{\mathbb{P}}[B]}} = {3\over 51}.\]

Note that this says that if the first card is an ace, it makes it less likely that the second card is also an ace.

# Bayesian Inference

Now that we’ve introduced the notion of conditional probability, we can see how it is used in real world settings. Conditional probability is at the heart of a subject called *Bayesian inference*, used extensively in fields such as machine learning, communications and signal processing. Bayesian inference is a way to *update knowledge* after making an observation. For example, we may have an estimate of the probability of a given event \(A\). After event \(B\) occurs, we can update this estimate to \({\mathbb{P}}[A \mid B]\). In this interpretation, \({\mathbb{P}}[A]\) can be thought of as a *prior* probability: our assessment of the likelihood of an event of interest \(A\) *before* making an observation. It reflects our prior knowledge. \({\mathbb{P}}[A \mid B]\) can be interpreted as the *posterior* probability of \(A\) after the observation. It reflects our new knowledge.

Here is an example of where we can apply such a technique. A pharmaceutical company is marketing a new test for a certain medical disorder. According to clinical trials, the test has the following properties:

When applied to an affected person, the test comes up positive in \(90\%\) of cases, and negative in \(10\%\) (these are called “false negatives”).

When applied to a healthy person, the test comes up negative in \(80\%\) of cases, and positive in \(20\%\) (these are called “false positives”).

Suppose that the incidence of the disorder in the US population is \(5\%\); this is our prior knowledge. When a random person is tested and the test comes up positive, how can we update this probability? (Note that this is presumably *not* the same as the simple probability that a random person has the disorder, which is just \(1/20\).) The implicit probability space here is the entire US population with uniform probabilities.

The sample space here consists of all people in the US — denote their number by \(N\) (so \(N\approx 250 \text{ million}\)). Let \(A\) be the event that a person chosen at random is affected, and \(B\) be the event that a person chosen at random tests positive. Now we can rewrite the information above:

\({\mathbb{P}}[A] = 0.05\) (\(5\%\) of the U.S. population is affected)

\({\mathbb{P}}[B \mid A] = 0.9\) (\(90\%\) of the affected people test positive)

\({\mathbb{P}}[B \mid \overline{A}] =0.2\) (\(20\%\) of healthy people test positive)

We want to calculate \({\mathbb{P}}[A \mid B]\). We can proceed as follows:

\[ {\mathbb{P}}[A \mid B] = \frac{{\mathbb{P}}[A\cap B]}{{\mathbb{P}}[B]} = \frac{{\mathbb{P}}[B \mid A]{\mathbb{P}}[A]}{{\mathbb{P}}[B]}\](1)

We obtained the second equality above by applying the definition of conditional probability: \[{\mathbb{P}}[B \mid A] = \frac{{\mathbb{P}}[A\cap B]}{{\mathbb{P}}[A]}\]

Now we need to compute \({\mathbb{P}}[B]\). This is the probability that a random person tests positive. To compute this, we can sum two values: the probability that a healthy person tests positive, \({\mathbb{P}}[\bar{A}\cap B]\) and the probability that an affected person tests positive, \({\mathbb{P}}[A\cap B]\). We can sum because the events \(\bar{A}\cap B\) and \(A\cap B\) do not intersect:

By again applying the definition of conditional probability we have: \[ {\mathbb{P}}[B] = {\mathbb{P}}[A \cap B] + {\mathbb{P}}[\overline{A} \cap B] = {\mathbb{P}}[B \mid A]{\mathbb{P}}[A] + {\mathbb{P}}[B \mid \overline{A}] (1-{\mathbb{P}}[A])\](2)

Combining Equation 1 and Equation 2, we have expressed \({\mathbb{P}}[A \mid B]\) in terms of \({\mathbb{P}}[A]\), \({\mathbb{P}}[B \mid A]\), and \({\mathbb{P}}[B \mid \overline{A}]\):

\[ {\mathbb{P}}[A \mid B] = \frac{{\mathbb{P}}[B \mid A] {\mathbb{P}}[A]}{{\mathbb{P}}[B \mid A]{\mathbb{P}}[A] + {\mathbb{P}}[B \mid \overline{A}] (1-{\mathbb{P}}[A])}\](3)

By plugging in the values written above, we obtain \({\mathbb{P}}[A \mid B] = 9/47 \approx .19\).

Equation 3 is useful for many inference problems. We are given \({\mathbb{P}}[A]\), which is the (unconditional) probability that the event of interest \(A\) happens. We are given \({\mathbb{P}}[B \mid A]\) and \({\mathbb{P}}[B \mid \overline{A}]\), which quantify how noisy the observation is. (If \({\mathbb{P}}[B \mid A] =1\) and \({\mathbb{P}}[B \mid \overline{A}] = 0\), for example, the observation is completely noiseless.) Now we want to calculate \({\mathbb{P}}[A \mid B]\), the probability that the event of interest happens given we made the observation. Equation 3 allows us to do just that.

Of course, Equation 1, Equation 2, and Equation 3 are derived from the basic axioms of probability and the definition of conditional probability, and are therefore true with or without the above Bayesian inference interpretation. However, this interpretation is very useful when we apply probability theory to study inference problems.

# Bayes Rule and Total Probability Rule

Equation 1 and Equation 2 are very useful in their own right. The first is called **Bayes Rule** and the second is called the **Total Probability Rule**. Bayes Rule is useful when one wants to calculate \({\mathbb{P}}[A \mid B]\) but one is given \({\mathbb{P}}[B \mid A]\) instead, i.e. it allows us to “flip” things around.

The Total Probability Rule is an application of the strategy of “dividing into cases” .There are two possibilities: either an event \(A\) happens or \(A\) does not happen. If \(A\) happens the probability that \(B\) happens is \({\mathbb{P}}[B \mid A]\). If \(A\) does not happen, the probability that \(B\) happens is \({\mathbb{P}}[B \mid \overline{A}]\). If we know or can easily calculate these two probabilities and also \({\mathbb{P}}[A]\), then the total probability rule yields the probability of event \(B\).

## Example: Tennis Match

You are about to play a tennis match against a randomly chosen opponent and you wish to calculate your probability of winning. You know your opponent will be one of two people, \(X\) or \(Y\). If person \(X\) is chosen, you will win with probability \(.7\). If person \(Y\) is chosen, you will win with probability \(.3\). Your opponent is chosen by flipping a coin with bias \(.6\) in favor of \(X\).

Let’s first determine which events we are interested in. Let \(A\) be the event that you win. Let \(B_1\) be the event that person \(X\) is chosen, and let \(B_2\) be the event that person \(Y\) is chosen. We wish to calculate \({\mathbb{P}}[A]\). Here is what we know so far:

\({\mathbb{P}}[A \mid B_1] = 0.7\), (if person \(X\) is chosen, you win with probability \(.7\))

\({\mathbb{P}}[A \mid B_2] = 0.3\) (if person \(Y\) is chosen, you win with probability \(.3\))

\({\mathbb{P}}[B_1] =0.6\) (person \(X\) is chosen with probability \(.6\))

\({\mathbb{P}}[B_2] =0.4\) (person \(Y\) is chosen with probability \(.4\))

By using the Total Probability rule, we have: \[{\mathbb{P}}[A] = {\mathbb{P}}[A \mid B_1]{\mathbb{P}}[B_1] + {\mathbb{P}}[A \mid B_2]{\mathbb{P}}[B_2].\]

Now we can simply plug in the known values above to obtain \({\mathbb{P}}[A]\): \[{\mathbb{P}}[A] = .7\times.6 + .3\times .4 = .54\]

## Example: Balls and Bins

Imagine we have two bins containing black and white balls, and further suppose that we wanted to know what is the chance that we picked bin \(1\) given that we picked a white ball, i.e., \({\mathbb{P}}[\text{Bin } 1 \mid \bigcirc]\). Assume that we are unbiased when choosing a bin so that each bin is chosen with probability \(1/2\).

A wrong approach is to say that the answer is clearly \(2/3\), since we know there are a total of three white balls, two of which are in bin \(1\). However, this picture is misleading because the bins have equal “weight". Instead, what we should do is appropriately scale each sample point as the following picture shows:

This images shows that the sample space \(\Omega\) is equal to the union of the events contained in bin \(1\) (\(A_1\)) and bin \(2\) (\(A_2\)), so \(\Omega = A_1\cup A_2\). We can use the definition of conditional probability to see that

\[{\mathbb{P}}[\text{Bin } 1 \mid \bigcirc] = \frac{1/10 + 1/10}{1/10 + 1/10 + 1/4} = \frac{2/10}{9/20} = \frac{4}{9}.\]

Let us try to achieve this probability using Bayes Rule. To apply Bayes Rule, we need to compute \({\mathbb{P}}[\bigcirc \mid \text{Bin } 1]\), \({\mathbb{P}}[\text{Bin } 1]\), and \({\mathbb{P}}[\bigcirc]\). \({\mathbb{P}}[\bigcirc \mid \text{Bin }1]\) is the chance that we pick a white ball given that we picked bin \(1\), which is \(2/5\). \({\mathbb{P}}[\text{Bin }1]\) is \(1/2\) as given in the description of the problem. Finally, \({\mathbb{P}}[\bigcirc]\) can be computed using the Total Probability Rule: \[{\mathbb{P}}[\bigcirc] = {\mathbb{P}}[\bigcirc \mid \text{Bin } 1] \cdot {\mathbb{P}}[\text{Bin } 1] + {\mathbb{P}}[\bigcirc \mid \text{Bin } 2] \cdot {\mathbb{P}}[\text{Bin } 2] = \frac{2}{5} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{9}{20}.\] Observe that we can apply the Total Probability Rule here because \({\mathbb{P}}[\text{Bin } 1]\) is the complement of \({\mathbb{P}}[\text{Bin } 2]\). Finally, if we plug the above values into Bayes Rule we obtain the probability that we picked bin \(1\) given that we picked a white ball: \[{\mathbb{P}}[\text{Bin } 1 \mid \bigcirc] = \frac{(2/5) \cdot (1/2)}{9/20} = \frac{2/10}{9/20} = \frac{4}{9}\] All we have done above is combined Bayes Rule and the Total Probability Rule; this is also how we obtained Equation 3. We could equivalently have plugged in the appropriate values to Equation 3.

# Combinations of Events

In most applications of probability in Computer Science, we are interested in things like \({\mathbb{P}}[\bigcup_{i=1}^n A_i]\) and \({\mathbb{P}}[\bigcap_{i=1}^n A_i]\), where the \(A_i\) are simple events (i.e., we know, or can easily compute, the \({\mathbb{P}}[A_i]\)). The intersection \(\bigcap_{i} A_i\) corresponds to the logical `AND`

of the events \(A_i\), while the union \(\bigcup_i A_i\) corresponds to their logical `OR`

. As an example, if \(A_i\) denotes the event that a failure of type \(i\) happens in a certain system, then \(\bigcup_i A_i\) is the event that the system fails.

In general, computing the probabilities of such combinations can be very difficult. In this section, we discuss some situations where it can be done. Let’s start with independent events, for which intersections are quite simple to compute.

## Independent Events

**Definition 2** *(Independence)*

*Two events \(A,B\) in the same probability space are independent if \({\mathbb{P}}[A\cap B]={\mathbb{P}}[A]\times {\mathbb{P}}[B]\).*

The intuition behind this definition is the following. Suppose that \({\mathbb{P}}[B]>0\). Then we have \[{\mathbb{P}}[A\mid B] = \frac{{\mathbb{P}}[A\cap B]}{{\mathbb{P}}[B]} = \frac{{\mathbb{P}}[A]\times {\mathbb{P}}[B]}{{\mathbb{P}}[B]} = {\mathbb{P}}[A].\] Thus independence has the natural meaning that “the probability of \(A\) is not affected by whether or not \(B\) occurs.” (By a symmetrical argument, we also have \({\mathbb{P}}[B\mid A] = {\mathbb{P}}[B]\) provided \({\mathbb{P}}[A]>0\).) For events \(A,B\) such that \({\mathbb{P}}[B]>0\), the condition \({\mathbb{P}}[A\mid B]={\mathbb{P}}[A]\) is actually *equivalent* to the definition of independence.

Several of our previously mentioned random experiments consist of independent events. For example, if we flip a coin twice, the event of obtaining heads in the first trial is independent to the event of obtaining heads in the second trial. The same applies for two rolls of a die; the outcomes of each trial are independent.

The above definition generalizes to any finite set of events:

**Definition 3** *(Mutual Independence)*

*Events \(A_1,\ldots,A_n\) are mutually independent if for every subset \(I\subseteq\{1,\ldots,n\}\),* \[{\textstyle{\mathbb{P}}[\bigcap_{i\in I}A_i] = \prod_{i\in I}{\mathbb{P}}[A_i]}.\]

Note that we need this property to hold for *every* subset \(I\).

For mutually independent events \(A_1,\ldots,A_n\), it is not hard to check from the definition of conditional probability that, for any \(1\le i\le n\) and any subset \(I\subseteq\{1,\ldots,n\}\setminus\{i\}\), we have \[{\textstyle{\mathbb{P}}[A_i\mid \bigcap_{j\in I}A_j] = {\mathbb{P}}[A_i]}.\]

Note that the independence of every pair of events (so-called *pairwise independence*) does *not* necessarily imply mutual independence. For example, it is possible to construct three events \(A,B,C\) such that each *pair* is independent but the triple \(A,B,C\) is *not* mutually independent.

### Pairwise Independence Example

Suppose you toss a fair coin twice and let \(A\) be the event that the first flip is \(H\) and \(B\) be the event that the second flip is \(H\). Now let \(C\) be the event that both flips are the same (i.e. both \(H\) or both \(T\)). Of course \(A\) and \(B\) are independent. What is more interesting is that so are \(A\) and \(C\): given that the first toss came up \(H\), there is still an even chance that the second flip is the same as the first. Another way of saying this is that \({\mathbb{P}}[A \cap C] = {\mathbb{P}}[A]{\mathbb{P}}[C] = 1/4\) since \(A \cap C\) is the event that the first flip is \(H\) and the second is also \(H\). By the same reasoning \(B\) and \(C\) are also independent. On the other hand, \(A\), \(B\), and \(C\) are not mutually independent. For example if we are given that \(A\) and \(B\) occurred then the probability that \(C\) occurs is \(1\). So even though \(A\), \(B\), and \(C\) are not mutually independent, every pair of them is independent. In other words, \(A\), \(B\), and \(C\) are pairwise independent but not mutually independent.

## Intersections of Events

Computing intersections of independent events is easy; it follows from the definition. We simply multiply the probabilities of each event. How do we compute intersections for events which may not be independent? From the definition of conditional probability, we immediately have the following *product rule* (sometimes also called the *chain rule*) for computing the probability of an intersection of events.

**Theorem 1** *(Product Rule)*

*For any events \(A,B\), we have \[{\mathbb{P}}[A\cap B]= {\mathbb{P}}[A]{\mathbb{P}}[B\mid A].\] More generally, for any events \(A_1,\ldots,A_n\), \[{\textstyle{\mathbb{P}}[\bigcap_{i=1}^n A_i] = {\mathbb{P}}[A_1]\times{\mathbb{P}}[A_2\mid A_1]\times
{\mathbb{P}}[A_3\mid A_1\cap A_2]\times
\cdots\times{\mathbb{P}}[A_n\mid\bigcap_{i=1}^{n-1}A_i]}.\]*

*Proof*. The first assertion follows directly from the definition of \({\mathbb{P}}[B\mid A]\) (and is in fact a special case of the second assertion with \(n=2\)).

To prove the second assertion, we will use induction on \(n\) (the number of events). The base case is \(n=1\), and corresponds to the statement that \({\mathbb{P}}[A]={\mathbb{P}}[A]\), which is trivially true. For the inductive step, let \(n>1\) and assume (the inductive hypothesis) that \[{\mathbb{P}}[{\textstyle\bigcap_{i=1}^{n-1} A_i}] = {\mathbb{P}}[A_1]\times{\mathbb{P}}[A_2\mid A_1]\times\cdots\times{\mathbb{P}}[A_{n-1}\mid{\textstyle\bigcap_{i=1}^{n-2} A_i}].\] Now we can apply the definition of conditional probability to the two events \(A_n\) and \(\bigcap_{i=1}^{n-1} A_i\) to deduce that \[\begin{aligned}{\mathbb{P}}[{\textstyle\bigcap_{i=1}^n A_i}] = {\mathbb{P}}[A_n\cap({\textstyle\bigcap_{i=1}^{n-1} A_i})] &= {\mathbb{P}}[A_n\mid{\textstyle\bigcap_{i=1}^{n-1} A_i}] \times{\mathbb{P}}[{\textstyle\bigcap_{i=1}^{n-1} A_i}] \\ &= {\mathbb{P}}[A_n\mid{\textstyle\bigcap_{i=1}^{n-1} A_i}] \times{\mathbb{P}}[A_1]\times{\mathbb{P}}[A_2\mid A_1]\times\cdots\times{\mathbb{P}}[A_{n-1}\mid{\textstyle\bigcap_{i=1}^{n-2} A_i}],\end{aligned}\] where in the last line we have used the inductive hypothesis. This completes the proof by induction. \(\square\)

The product rule is particularly useful when we can view our sample space as a sequence of choices. The next few examples illustrate this point.

## Examples

### Coin Tosses

Toss a fair coin three times. Let \(A\) be the event that all three tosses are heads. Then \(A=A_1\cap A_2\cap A_3\), where \(A_i\) is the event that the \(i\)th toss comes up heads. We have \[\begin{aligned} {\mathbb{P}}[A]&={\mathbb{P}}[A_1]\times{\mathbb{P}}[A_2\mid A_1]\times{\mathbb{P}}[A_3\mid A_1\cap A_2] \\ &={\mathbb{P}}[A_1]\times{\mathbb{P}}[A_2]\times{\mathbb{P}}[A_3] \\ &=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} ={1\over 8}. \end{aligned}\] The second line here follows from the fact that the tosses are mutually independent. Of course, we already know that \({\mathbb{P}}[A]=1/8\) from our definition of the probability space in the previous lecture note. Another way of looking at this calculation is that it justifies our definition of the probability space, and shows that it was consistent with assuming that the coin flips are mutually independent.

If the coin is biased with heads probability \(p\), we get, again using independence, \[{\mathbb{P}}[A] = {\mathbb{P}}[A_1]\times {\mathbb{P}}[A_2]\times {\mathbb{P}}[A_3] = p^3.\] And more generally, the probability of any sequence of \(n\) tosses containing \(r\) heads and \(n-r\) tails is \(p^r(1-p)^{n-r}\). This is in fact the reason we defined the probability space this way in the previous lecture note: we defined the sample point probabilities so that the coin tosses would behave independently.

### Monty Hall

Recall the Monty Hall problem from the last lecture: there are three doors and the probability that the prize is behind any given door is \(1/3\). There are goats behind the other two doors. The contestant picks a door randomly, and the host opens one of the other two doors, revealing a goat. How do we calculate intersections in this setting? For example, what is the probability that the contestant chooses door \(1\), the prize is behind door \(2\), and the host chooses door \(3\)?

Let \(A_1\) be the event that the contestant chooses door \(1\), let \(A_2\) be the event that the prize is behind door \(2\), and let \(A_3\) be the event that the host chooses door \(3\). We would like to compute \({\mathbb{P}}[A_1\cap A_2\cap A_3]\). By the product rule: \[{\mathbb{P}}[A_1\cap A_2\cap A_3] = {\mathbb{P}}[A_1]\times {\mathbb{P}}[A_2 \mid A_1] \times {\mathbb{P}}[A_3 \mid A_1\cap A_2]\]

The probability of \(A_1\) is \(1/3\), since the contestant is choosing the door at random. The probability \(A_2\) given \(A_1\) is still \(1/3\) since they are independent. The probability of the host choosing door \(3\) given events \(A_1\) and \(A_2\) is \(1\); the host cannot choose door \(1\), since the contestant has already opened it, and the host cannot choose door \(2\), since the host must reveal a goat (and not the prize). Therefore, \[{\mathbb{P}}[A_1\cap A_2\cap A_3] = \frac{1}{3}\times\frac{1}{3}\times 1 = \frac{1}{9}.\]

Observe that we did need conditional probability in this setting; had we simply multiplied the probabilities of each event, we would have obtained \(1/27\) since the probability of \(A_3\) is also \(1/3\) (can you figure out why?). What if we changed the situation, and instead asked for the probability that the contestant chooses door \(1\), the prize is behind door \(1\), and the host chooses door \(2\)? We can use the same technique as above, but our final answer will be different. This is left as an exercise.

### Poker Hands

Let’s use the product rule to compute the probability of a flush in a different way. This is equal to \(4\times{\mathbb{P}}[A]\), where \(A\) is the probability of a Hearts flush. Intuitively, this should be clear since there are \(4\) suits; we’ll see why this is formally true in the next section. We can write \(A=\bigcap_{i=1}^5 A_i\), where \(A_i\) is the event that the \(i\)th card we pick is a Heart. So we have \[{\textstyle{\mathbb{P}}[A]={\mathbb{P}}[A_1]\times{\mathbb{P}}[A_2\mid A_1]\times\cdots\times {\mathbb{P}}[A_5\mid\bigcap_{i=1}^4 A_i]}.\] Clearly \({\mathbb{P}}[A_1]=13/52 = 1/4\). What about \({\mathbb{P}}[A_2\mid A_1]\)? Well, since we are conditioning on \(A_1\) (the first card is a Heart), there are only \(51\) remaining possibilities for the second card, \(12\) of which are Hearts. So \({\mathbb{P}}[A_2\mid A_1]=12/51\). Similarly, \({\mathbb{P}}[A_3\mid A_1\cap A_2]=11/50\), and so on. So we get \[4\times{\mathbb{P}}[A]=4\times{{13}\over{52}}\times{{12}\over{51}}\times{{11}\over{50}}\times{{10}\over{49}}\times{{9}\over{48}},\] which is exactly the same fraction we computed in the previous lecture note.

So now we have two methods of computing probabilities in many of our sample spaces. It is useful to keep these different methods around, both as a check on your answers and because in some cases one of the methods is easier to use than the other.

## Unions of Events

You are in Las Vegas, and you spy a new game with the following rules. You pick a number between \(1\) and \(6\). Then three dice are thrown. You win if and only if your number comes up on at least one of the dice.

The casino claims that your odds of winning are \(50\%\), using the following argument. Let \(A\) be the event that you win. We can write \(A=A_1\cup A_2\cup A_3\), where \(A_i\) is the event that your number comes up on die \(i\). Clearly \({\mathbb{P}}[A_i]=1/6\) for each \(i\). Therefore, \[{\mathbb{P}}[A]={\mathbb{P}}[A_1\cup A_2\cup A_3] = {\mathbb{P}}[A_1]+{\mathbb{P}}[A_2]+{\mathbb{P}}[A_3] = 3\times{1\over 6} = {1\over 2}.\] Is this calculation correct? Well, suppose instead that the casino rolled six dice, and again you win iff your number comes up at least once. Then the analogous calculation would say that you win with probability \(6\times (1/6)=1\), i.e., certainly! The situation becomes even more ridiculous when the number of dice gets bigger than \(6\).

The problem is that the events \(A_i\) are *not disjoint*: i.e., there are some sample points that lie in more than one of the \(A_i\). (We could get really lucky and our number could come up on two of the dice, or all three.) So if we add up the \({\mathbb{P}}[A_i]\) we are counting some sample points more than once.

Fortunately, there is a formula for this, known as the *Principle of Inclusion/Exclusion*:

**Theorem 2** *(Inclusion/Exclusion)*

*For events \(A_1,\ldots,A_n\) in some probability space, we have \[{\mathbb{P}}[{\textstyle\bigcup_{i=1}^n A_i}] = \sum_{i=1}^n {\mathbb{P}}[A_i]
-\sum_{\{i,j\}} {\mathbb{P}}[A_i\cap A_j]
+\sum_{\{i,j,k\}} {\mathbb{P}}[A_i\cap A_j\cap A_k]
-\cdots\pm {\mathbb{P}}[{\textstyle\bigcap_{i=1}^n A_i}].\] [In the above summations, \(\{i,j\}\) denotes all unordered pairs with \(i\ne j\), \(\{i,j,k\}\) denotes all unordered triples of distinct elements, and so on.]*

In other words, to compute \({\mathbb{P}}[\bigcup_i A_i]\), we start by summing the event probabilities \({\mathbb{P}}[A_i]\), then we *subtract* the probabilities of all pairwise intersections, then we *add* back in the probabilities of all three-way intersections, and so on.

We won’t prove this formula here; but you might like to verify it for the special case \(n=3\) by drawing a Venn diagram and checking that every sample point in \(A_1\cup A_2\cup A_3\) is counted exactly once by the formula. You might also like to prove the formula for general \(n\) by induction (in similar fashion to the proof of the Product Rule above).

Taking the formula on faith, what is the probability we get lucky in the new game in Vegas? \[{\mathbb{P}}[A_1\cup A_2\cup A_3] = {\mathbb{P}}[A_1]+{\mathbb{P}}[A_2]+{\mathbb{P}}[A_3] -{\mathbb{P}}[A_1\cap A_2] -{\mathbb{P}}[A_1\cap A_3] -{\mathbb{P}}[A_2\cap A_3] +{\mathbb{P}}[A_1\cap A_2\cap A_3].\] Now the nice thing here is that the events \(A_i\) are mutually independent (the outcome of any die does not depend on that of the others), so \({\mathbb{P}}[A_i\cap A_j]={\mathbb{P}}[A_i]{\mathbb{P}}[A_j] = (1/36)^2=1/36\), and similarly \({\mathbb{P}}[A_1\cap A_2\cap A_3]=(1/6)^3=1/216\). So we get \[{\mathbb{P}}[A_1\cup A_2\cup A_3] = \left(3\times{1\over 6}\right) - \left(3\times{1\over {36}}\right) + {1\over{216}} = {{91}\over{216}}\approx 0.42.\] So your odds are quite a bit worse than the casino is claiming!

When \(n\) is large (i.e., we are interested in the union of many events), the Inclusion/Exclusion formula is essentially useless because it involves computing the probability of the intersection of every non-empty subset of the events: and there are \(2^n-1\) of these! Sometimes we can just look at the first few terms of it and forget the rest: note that successive terms actually give us an overestimate and then an underestimate of the answer, and these estimates both get better as we go along.

However, in many situations we can get a long way by just looking at the term:

**Disjoint events.**If the events \(A_i\) are all*disjoint*(i.e., no pair of them contain a common sample point — such events are also called*mutually exclusive*), then \[{\mathbb{P}}[{\textstyle\bigcup_{i=1}^n A_i}] = \sum_{i=1}^n {\mathbb{P}}[A_i].\] [Note that we have already used this fact several times in our examples, e.g., in claiming that the probability of a flush is four times the probability of a Hearts flush — clearly flushes in different suits are disjoint events.]**Union bound.**Always, it is the case that \[{\mathbb{P}}[{\textstyle\bigcup_{i=1}^n A_i}] \le \sum_{i=1}^n {\mathbb{P}}[A_i].\] This merely says that adding up the \({\mathbb{P}}[A_i]\) can only*over*estimate the probability of the union. Crude as it may seem, in the next lecture note we’ll see how to use the union bound effectively in a Computer Science example.