Event operations

• \(E\) is a set, or a collection of distinct objects. We can perform operations on sets to create new sets. Consider two example sets:

• \(E = \{ (W,W), (W,N) \}\)
• \(F = \{ (N, N), (W,N) \}\)

• \(\Omega = \{(W,W), (W,N), (N,W), (N,N) \}\)

Operations determine what lies in the new set \(E^{\text{new}}\).

1. Union: \(\cup\)
   • All objects that appear in either set (OR)
   • \(E^{\text{new}} = E \cup F = \{(W,W), (W,N), (N,N) \}\)
2. Intersection: \(\cap\)
   • All objects that appear in both sets (AND)
   • \(E^{\text{new}} = E \cap F = \{(W,N)\}\)
3. Complement of set \(E\): \(E^{c}\)
   • All objects in \(S\) that are not in \(E\)
   • \(E^{c} = \{(N, W) , (N, N) \}\)
   • \(F^{c} = \{(N, W) , (W, W) \}\)
   • What is \(\Omega^{c}\)? - an empty set \(\emptyset\)
   • Suppose \(E = {W}\), \(F = {N}\). Then \(E \cap F = \emptyset\) (there is nothing that lies in both sets)

Three axioms

All probability functions \(P\) satisfy three axioms:

1. Nonnegativity: For all events \(E\), \(0 \leq P(E) \leq 1\)
2. Normalization: \(P(S) = 1\)
3. Additivity: For all sequences of mutually exclusive events \(E_{1}, E_{2}, \ldots,E_{N}\) (where \(N\) can go to infinity): \[P\left(\cup_{i=1}^{N} E_{i} \right) = \sum_{i=1}^{N} P(E_{i} )\]
   • Any countable sequence of mutually exclusive events can be added together to generate the probability of any of the mutually exclusive events occurring

Basic examples

Rolling the dice

Consider the experiment of rolling a pair of 4-sided dice. We assume the dice are fair, and we interpret this assumption to mean that each of the sixteen possible outcomes [pairs \((i,j)\) with \(i,j = 1,2,3,4\)] has the same probability of \(1/16\). To calculate the probablity of an event, we must count the number of elements of the event and divide by 16 (the total number of possible outcomes). Here are some event probabilities calculated this way:

\[ \begin{aligned} \Omega &= \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), \\ &\quad (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4) \} \end{aligned} \]

\[ \begin{aligned} \Pr (\text{the sum of the rolls is even}) &= 8/16 = 1/2 \\ \Pr (\text{the sum of the rolls is odd}) &= 8/16 = 1/2 \\ \Pr (\text{the first roll is equal to the second}) &= 4/16 = 1/4 \\ \Pr (\text{the first roll is larger than the second}) &= 6/16 = 3/8 \\ \Pr (\text{at least one roll is equal to 4}) &= 7/16 \end{aligned} \]

Romeo and Juliet

Romeo and Juliet have a date at a given time, and each will arrive at the meeting place with a delay between 0 and 1 hour, with all pairs of delays being equally likely. The first to arrive will wait for 15 minutes and will leave if the other has not yet arrived. What is the probability that they will meet?

data_frame(x = seq(from = 0, to = 1, by = 0.001),
           ylow = -.25 + x,
           yhigh = .25 + x) %>%
  ggplot(aes(x = x)) +
  geom_line(aes(y = ylow)) +
  geom_line(aes(y = yhigh)) +
  geom_ribbon(aes(ymin = ylow, ymax = yhigh), alpha = .2) +
  annotate(geom = "text", x = .5, y = .5, label = "M") +
  scale_x_continuous(breaks = c(0, .25, 1)) +
  scale_y_continuous(breaks = c(0, .25, 1)) +
  coord_fixed(xlim = c(0, 1),
              ylim = c(0, 1),
              expand = FALSE) +
  labs(x = "Romeo",
       y = "Juliet") +
  theme_classic() +
  theme(panel.border = element_rect(color = "black", fill = NA, size = 1))

Examples

Design teams

A conservative design team, call it \(C\), and an innovative design team, call it \(N\), are asked to separately design a new product within a month. From past experience we know:

1. The probability that team \(C\) is successful is \(2/3\).
2. The probability that team \(N\) is successful is \(1/2\).
3. The probability that at least one team is successful is \(3/4\).

Assuming that exactly one successful design is produced, what is the probability that it was designed by team \(N\)?

Difference between \(P(A|B)\) and \(P(B|A)\)

\[ \begin{eqnarray} P(A|B) & = & \frac{P(A\cap B)}{P(B)} \\ P(B|A) & = & \frac{P(A \cap B) } {P(A)} \end{eqnarray} \]

Less Serious Example \(\leadsto\) type of person who attends football games:

\[ \begin{eqnarray} P(\text{Attending a football game}| \text{Drunk}) & = & 0.01 \\ P(\text{Drunk}| \text{Attending a football game}) & \approx & 1 \end{eqnarray} \]

Example: chess tournament

You enter a chess tournament where your probability of winning a game is \(0.3\) against half the players (type 1), \(0.4\) against a quarter of the players (type 2), and \(0.5\) against the remaining quarter of the players (type 3). You play a game against a randomly chosen opponent. What is the probability of winning?

Examples

Examples

Independence of a collection of events

We say that the events \(A_1, A_2, \ldots, A_n\) are independent if

\[ \Pr \left( \bigcap_{i \in S} A_i \right) = \prod_{i \in S} \Pr (A_i),\quad \text{for every subset } S \text{ of } \{1,2,\ldots,n \}\]

For the case of three events, \(A_1, A_2, A_3\), independence amounts to satisfying the four conditions

\[ \begin{aligned} \Pr (A_1 \cap A_2) &= \Pr (A_1) \Pr (A_2) \\ \Pr (A_1 \cap A_3) &= \Pr (A_1) \Pr (A_3) \\ \Pr (A_2 \cap A_3) &= \Pr (A_2) \Pr (A_3) \\ \Pr (A_1 \cap A_2 \cap A_3) &= \Pr (A_1) \Pr (A_2) \Pr (A_3) \end{aligned} \]

Independent trials and the binomial probabilities

If an experiment involves a sequence of independent but identical stages, we say that we have a sequence of independent trials. In the case where there are only two possible results of each stage, we say that we have a sequence of independent Bernoulli trials.

• Heads or tails
• Success or fail
• Rains or does not rain

Consider an experiment that consists of \(n\) independent tosses of a coin, in which the probability of heads is \(p\), where \(p\) is some number between 0 and 1. In this context, independence means that the events \(A_1, A_2, \ldots, A_n\) are independent where \(A_i = i \text{th toss is a heads}\).

Let us consider the probability

\[p(k) = \Pr(k \text{ heads come up in an } n \text{-toss sequence})\]

The probability of any given sequence that contains \(k\) heads is \(p^k (1-p)^{n-k}\), so we have

\[p(k) = \binom{n}{k} p^k (1-p)^{n-k}\]

where we use the notation

\[\binom{n}{k} = \text{number of distinct } n \text{-toss sequences that contain } k \text{ heads}\]

The numbers \(\binom{n}{k}\) (read as “\(n\) choose \(k\)”) are known as the binomial coefficients, while the probabilities \(p(k)\) are known as the binomial probabilities. Using a counting argument, we can show that

\[\binom{n}{k} = \frac{n!}{k! (n-k)!}, \quad k=0,1,\ldots,n\]

where for any positive integer \(i\) we have

\[i! = 1 \times 2 \times \cdots \times (i-1) \times i\]

and, by convention, \(0! = 1\). Note that the binomial probabilities \(p(k)\) must sum to 1, thus showing the binomial formula

\[\sum_{k=0}^n \binom{n}{k} p^k (1-p)^{n-k} = 1\]

Examples

Counting principle

Consider a process that consists of \(r\) stages. Suppose that:

1. There are \(n_1\) possible results at the first stage.
2. For every possible result at the first stage, there are \(n_2\) possible results at the second stage.
3. More generally, for any sequence of possible results at the first \(i-1\) stages, there are \(n_i\) possible results at the \(i\)th stage.

Then, the total number of possible results of the \(r\)-stage process is

\[n_1, n_2, \cdots, n_r\]

Permutations

We start with \(n\) distinct objects, and let \(k\) be some positive integer such that \(k \leq n\). We wish to count the number of different ways that we can pick \(k\) out of these \(n\) objects and arrange them in a sequence (i.e. the number of distinct \(k\)-object sequences). The number of possible sequences, called \(k\)-permutations, is

\[ \begin{aligned} n(n-1) \cdots (n-k-1) &= \frac{n(n-1) \cdots (n-k+1) (n-k) \cdots 2 \times 1}{(n-k) \cdots 2 \times 1} \\ &= \frac{n!}{(n-k)!} \end{aligned} \]

Combinations

There are \(n\) people and we are interested in forming a committee of \(k\). How many different committees are possible? Notice that this is a counting problem inside of a counting problem: we need to count the number of \(k\)-element subsets of a given \(n\)-element set. In a combination, there is no ordering of the selected elements. For example, whereas the 2-permutations of the letters \(A, B, C, D\) are

\[AB, BA, AC, CA, AD, DA, BC, CB, BD, DB, CD, DC\]

the combinations of two out of these four letters are

\[AB, AC, AD, BC, BD, CD\]

In this example, we group together duplicates that are not distinct and tabulate their frequency. More generally, we can view each combination as associated with \(k!\) duplicate \(k\)-permutation. Hence, the number of possible combinations is equal to

\[\frac{n!}{k!(n-k)!}\]