## 2.2 Events

In a random phenomenon, an “event” is something that could happen. For example, if we’re interested in tomorrow’s weather, events include

- “the high temperature is 75°F”
- “the high temperature is at least 75°F”
- “it rains”
- “it does not rain”
- “it rains and the high temperature is 75°F”
- and so on

Mathematically, an *event* is a collection of sample space outcomes that satisfy some criteria.

**Definition 2.2**An

**event**\(A\) is a

*subset*of the sample space: \(A\subseteq \Omega\). The collection of all events of interest

^{10}is denoted \(\mathcal{F}\). If the random phenomenon yields outcome \(\omega\), we say “event \(A\) occurred” if \(\omega\in A\).

- Recall that the sample space is the collection of all possible outcomes of a random phenomenon.
- An event represents those outcomes which satisfy some criteria.
- There is a single sample space, upon which all events are defined.
- An outcome in a sample space should be defined to record as much information as possible so that the occurrence or non-occurrence of all events of interest can be determined.
- Events are typically denoted with capital letters near the start of the alphabet, with or without subscripts (e.g. \(A\), \(B\), \(C\), \(A_1\), \(A_2\))
- Events can be composed from others using basic set operations like unions (\(A\cup B\)), intersections (\(A \cap B\)), and complements (\(A^c\)).
- Read \(A^c\) as “not” \(A\).
- Read \(A\cap B\) as “\(A\) and \(B\)”
- Read \(A \cup B\) as “\(A\) or \(B\)”. Note that unions (\(\cup\), “or”) are always inclusive. \(A\cup B\) occurs if \(A\) occurs but \(B\) does not, \(B\) occurs but \(A\) does not, or both \(A\) and \(B\) occur.

- An event \(A\) is a set. The collection \(\mathcal{F}\) is a
*collection of sets*. - While an event is always a set, it can be a set consisting of a single outcome, or no outcomes at all (the empty set \(\emptyset\)).
- A collection of events \(A_1, A_2, \ldots\) are
**disjoint**(a.k.a. mutually exclusive) if \(A_i \cap A_j = \emptyset\) for all \(i \neq j\). Events are disjoint if they have no outcomes in common.

As an example, consider a single roll of a four-sided die. The sample space consists of the four possible outcomes \(\Omega = \{1, 2, 3, 4\}\). Any subset of this sample space is an event. The following table lists he collection of all events (\(\mathcal{F}\)), and whether they occur if the single roll results in a 3 (that is, for the outcome \(\omega=3\)).

Event | Description | Occurs upon outcome \(\omega=3\)? |
---|---|---|

\(\emptyset\) | Roll nothing (not possible) | No |

\(\{1\}\) | Roll a 1 | No |

\(\{2\}\) | Roll a 2 | No |

\(\{3\}\) | Roll a 3 | Yes |

\(\{4\}\) | Roll a 4 | No |

\(\{1, 2\}\) | Roll a 1 or a 2 | No |

\(\{1, 3\}\) | Roll a 1 or a 3 | Yes |

\(\{1, 4\}\) | Roll a 1 or a 4 | No |

\(\{2, 3\}\) | Roll a 2 or a 3 | Yes |

\(\{2, 4\}\) | Roll a 2 or a 4 | No |

\(\{3, 4\}\) | Roll a 3 or a 4 | Yes |

\(\{1, 2, 3\}\) | Roll a 1, 2, or 3 (a.k.a. do not roll a 4) | Yes |

\(\{1, 2, 4\}\) | Roll a 1, 2, or 4 (a.k.a. do not roll a 3) | No |

\(\{1, 3, 4\}\) | Roll a 1, 3, or 4 (a.k.a. do not roll a 2) | Yes |

\(\{2, 3, 4\}\) | Roll a 2, 3, or 4 (a.k.a. do not roll a 1) | Yes |

\(\{1, 2, 3, 4\}\) | Roll something | Yes |

**Example 2.8**Roll a four-sided die

*twice*, and record the result of each roll in sequence. For example, the outcome \((3, 1)\) represents a 3 on the first roll and a 1 on the second; this is not the same outcome as \((1, 3)\).

- Identify the sample space.
- Identify \(A\), the event that the sum of the two dice is 4.
- Identify \(B\), the event that the sum of the two dice is at most 3.
- The previous events consider the sum of the two dice. Explain why we don’t just consider the sample space to be \(\{2, \ldots, 8\}\), so that for example \(B = \{2, 3\}\).
- Identify \(C\), the event that the larger of the two rolls (or the common roll if a tie) is 3
- Identify and interpret \(A\cap C\).
- Identify \(D\), the event that the first roll is a 3.
- Identify \(E\), the event that the second roll is a 3.
- Indetify and interpret \(D \cap E\).
- Identify and interpret \(D \cup E\).
- If the outcome is \((1, 3)\), which of the events above occurred?

*Solution*to Example 2.8

- The sample space consists of 16 possible ordered pairs of rolls \[\begin{align*} \Omega & = \{(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4),\\ & \quad (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)\} \end{align*}\]
- \(A=\{(1, 3), (2, 2), (3, 1)\}\) is the event that the sum of the two dice is 4.
- \(B=\{(1, 1), (1, 2), (2, 1)\}\) is the event that the sum of the two dice is at most 3.
- We might be interested in events other than ones that involve the sum of the dice, like those in the following parts. Knowing just the sum of the dice does not provide enough information to investigate such events.
- \(C=\{(1, 3), (2, 3), (3, 1), (3, 2), (3, 3)\}\), the event that the larger of the two rolls (or the common roll if a tie) is 3
- \(A\cap C=\{(1, 3), (3, 1)\}\) is the event that both the sum of the two dice is 4 and the larger of the two rolls is 3.
- \(D=\{(3, 1), (3, 2), (3, 3), (3, 4)\}\), the event that the first roll is a 3. Note that each outcome in the sample space consists of a pair of rolls, so we must account for both rolls in defining events, even if the event of interest involves just the first roll. There is always a single sample space upon which all events are defined.
- \(E=\{(1, 3), (2, 3), (3, 3), (4, 3)\}\), the event that the second roll is a 3. Note that this is not the same event as \(D\).
- \(D \cap E = \{(3, 3)\}\) is the event that both rolls result in a 3. While an event is always a set, it can be a set consisting of a single outcome (or the empty set).
- \(D \cup E = \{(3, 1), (3, 2), (3, 3), (3, 4), (1, 3), (2, 3), (4, 3)\}\) is the event that at least one of the two rolls results in a 3. Notice that the union is inclusive: \((3, 3)\) is an element of \(D\cup E\). But also notice that the outcome \((3, 3)\) only appears once in the set \(D \cup E\).
- If the outcome is \((1, 3)\) then events \(A\), \(C\), \(A\cap C\), \(E\), \(D\cup E\) all occur. Events \(B\), \(D\), and \(D\cap E\) do not occur.

**Example 2.9**Flip a coin 4 times, and record the result of each trial in sequence. For example, HTTH means heads on the first on last trial and tails on the second and third.

- Specify the sample space.
- Identify \(A\), the event that exactly 3 of the flips land on heads.
- Identify \(B\), the event that exactly 4 of the flips land on heads.
- Identify \(C\), the event that the at least 3 of the flips land on heads. How does \(C\) relate to \(A\) and \(B\)?
- The previous events all consider the number of heads flipped. Explain why we don’t just consider the sample space to be \(\{0, 1, 2, 3, 4\}\), so that for example \(C = \{3, 4\}\).
- Identify \(D\), the event that at least 3 heads are flipped in a row.
- Identify \(E\), the event that the first two flips results in tails.
- Identify \(D\cap E\).
- If the outcome is HHTH, which of the events above occurred?

*Solution*to Example 2.9

- The sample space is composed of 16 outcomes \[\begin{align*} \Omega & = \{HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH,\\ & \qquad THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT\} \end{align*}\]
- \(A = \{HHHT, HHTH, HTHH, THHH\}\) is the event that exactly 3 of the flips land on heads
- \(B = \{HHHH\}\), the event that exactly 4 of the flips land on heads. While an event is always a set, it can be a set consisting of a single outcome.
- \(C = \{HHHT, HHTH, HTHH, THHH, HHHH\}\) is the event that the at least 3 of the flips land on heads. Also \(C = A \cup B\).
- Yes, the previous events all consider the number of heads flipped, but we might be interested in events — like the ones in the following parts — whose occurrence cannot be determined simply by knowing the number of heads. The sample space should always to defined in such a way to provide enough information to investigate any relevant event of interest. There is always a single sample space upon which all events are defined.
- \(D = \{HHHT, THHH, HHHH\}\) is the event that at least 3 heads are flipped in a row. Note that \(D\) is not the same event as \(C\).
- \(E = \{TTHH, TTHT, TTTH, TTTT\}\) is the event that the first two flips result in tails. Note that each outcome consists of the results of 4 flips, so we must account for all 4 flips in defining events. There is always a single sample space upon which all events are defined.
- \(D\cap E=\emptyset\) so the events \(D\) and \(E\) are disjoint; it is not possible to have at least three heads in a row when the first two flips (out of 4) are tails.
- If the outcome is HHTH then events \(A\) and \(C\) occur. Events \(B\), \(D\), \(E\), and \(D \cap E\) do not occur.

**Example 2.10 (Matching problem)**So-called “matching problems” concern the following generic scenario. A set of \(n\) cards labeled \(1, 2, \ldots, n\) are placed in \(n\) boxes labeled \(1, 2, \ldots, n\), with exactly one card in each box. Typical questions of interest involve whether the number of a card matches the number of the box in which it is placed. (More colorful descriptions include returning babies at random to mothers or placing rocks at random back on a museum shelf.) Consider the case \(n=4\).

- Identify a reasonable sample space.
- Specify the event that all cards match their box.
- Specify the event that no cards match their box.
- Specify the event that exactly 3 cards match their box.
- Specify the event that card 3 is placed in box 3.

*Solution*to Example 2.10

- We can consider each outcome to be a particular placement of cards in the boxes. For example, the outcome 3214 (or \((3, 2, 1, 4)\)) represents that card 3 is placed in box 1, card 2 in box 2, card 1 in box 3, and card 4 in box 4. So the sample space consists of the following 24 outcomes
^{11}. \[\begin{align*} \Omega & = \{1234, 1243, 1324, 1342, 1423, 1432 \\ & \qquad 2134, 2143, 2314, 2341, 2413, 2431 \\ & \qquad 3124, 3142, 3214, 3241, 3412, 3421 \\ & \qquad 4123, 4132, 4213, 4231, 4312, 4321\} \end{align*}\] - \(A=\{1234\}\) is the event that all cards match their box. While an event is always a set, it can be a set consisting of a single outcome (or the empty set).
- \(B=\{2143, 2341, 2413, 3142, 3412, 3421, 4123, 4312, 4321\}\) is the event that no cards match their box.
- \(\emptyset\) is the event in which exactly 3 cards match their box. There are no outcomes in which exactly 3 cards match; if three cards match, then the fourth card must necessarily match too.
- \(C=\{1234, 1432, 2134, 2431, 4132, 4231\}\) is the event that card 3 is placed in box 3. Since our sample space consists of the placements of each of the cards, each event must be expressed in terms of these outcomes.

Remember that intervals of real numbers such as \((a,b), [a,b], (a,b]\) are also sets, and so can also be events.

**Example 2.11**Consider as a sample space all possible scores on the SAT Math exam (ranging from 200 to 800). Even though actual scores are multiples of 10, it is mathematically convenient to consider the sample space as \(\Omega=[200, 800]\), the set of all real numbers between 200 and 800 (rather than \(\{200, 210, 220, \ldots, 800\}\).)

- Identify \(A\), the event that an SAT Math score is at most 700.
- Identify \(B\), the event that an SAT Math score is greater than 500.
- Identify and interpret \(A\cap B\).

*Solution*to Example 2.11

- \(A=[200, 700]\) is the event that an SAT Math score is at most 700.
- \(B=(500, 800]\) is the event that an SAT Math score is greater than 500.
- \(A\cap B = (500, 700]\) is the event that an SAT Math score is greater than 500 but at most 700.

**Example 2.12 **
Suppose the spinner in Figure 2.1 is spun twice. Let the sample space be \(\Omega = [0,1]\times [0,1]\). Represent each of the following events using set notation, but also sketch a picture. (Hint: represent the sample space as a square.)

- Identify \(A\), the event that the first spin is larger then the second.
- Identify \(B\), the event that the smaller of the two spins (or common value if a tie) is less than 0.5.
- Identify \(C\), the event that the sum of the two dice is less than 1.5.
- Identify \(D\), the event that the first spin is less than 0.4.

*Solution *
to Example 2.12

- \(A = \{(\omega_1, \omega_2): \omega_1>\omega_2\}\) is the event that the first spin is larger then the second. Note: we only consider \((\omega_1, \omega_2)\) in the sample space \([0, 1]\times[0,1]\); the conditions \(0\le \omega_1 \le 1, 0\le \omega_2 \le 1\) are assumed.

- \(B = \{(\omega_1, \omega_2): \min(\omega_1,\omega_2)<0.5\}\) is the event that the smaller of the two spins (or common value if a tie) is less than 0.5. This is equivalent to the event that at least one of the spins is less than 0.5 \(B = \{(\omega_1, \omega_2): \omega_1<0.5 \text{ or } \omega_2<0.5\} = ([0.5, 1]\times[0.5, 1])^c\).
- \(C = \{(\omega_1, \omega_2): \omega_1+\omega_2\ge 1.5\} = \{(\omega_1, \omega_2): \omega_2\ge 1.5-\omega_1\}\) is the event that the sum of the two dice is at least 1.5.
- \(D = \{(\omega_1, \omega_2): \omega_1<0.4\}\) is the event that the first spin is less than 0.4. Remember, sample space outcomes are pairs of spins.

**Example 2.13 (Don’t do what Donny Don’t does.)**Donny Don’t is asked a series of questions involving a pair of rolls of six-sided dice, such as “what is the event that the sum of the dice is at least 10”. Donny’s responses are below; explain to him what is wrong with his responses and help him see the correct answers.

- The possible rolls are 1 through 6, so the sample space is \(\{1, 2, 3, 4, 5, 6\}\).
- The sum of the two dice can be 2 through 12, so the event that the sum of the two dice is at least 10 is \(\{10, 11, 12\}\).
- The event that the first roll is a 3 is \(\{3\}\).
- The event that the first roll is a 3 and the second roll is a 1 is \(\{3, 1\}\)
- Donny’s sample space from the first question might correspond to what dice rolling scenario? What does \(\{3, 1\}\) represent in this scenario?

*Solution*to Example 2.13

- The questions involve a pair of rolls, so best to record an outcome as an ordered pair, e.g., (5, 2) for 5 on the first roll and 2 on the second. Therefore, the sample space would be the following set of 36 possible outcomes. \[\begin{align*} \Omega = & \{ (1, 1), (1, 2), \ldots, (1, 6),\\ & \;\; (2, 1), (2, 2), \ldots, (2, 6),\\ & \;\; \vdots\qquad \qquad \quad \cdots \qquad \vdots\\ & \;\; (6, 1), (6, 2), \ldots, (6, 6) \} \end{align*}\]
- Donny’s answers to the first two parts are inconsistent, since there is always a single sample space. So if he says the answer to the first part is \(\{1, \ldots, 6\}\), then any event must be a subset of that sample space and his answer to the second part must be wrong. Using the sample space of 36 ordered pairs from the previous answer, the correct event that the sum of the two dice is at least 10 is \[ \{(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)\} \] If Donny’s sample space in the first part had been \(\{2, \ldots, 12\}\), corresponding to the sum of the two dice, then his answer of \(\{10, 11, 12\}\) would have been correct. However, using such a sample space, he would not have been able to answer the remaining questions (which don’t involve the sum of the rolls). There is always one sample space on which all events are defined.
- Donny didn’t take into account that an outcome is a pair of rolls. The correct event is \(\{(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)\}\), the set of all pairs of rolls for which the first roll is 3.
- Maybe Donny is just using bad notation here, but it sure looks like he is confusing an outcome with an event. The answer should be \(\{(3, 1)\}\), the set containing the single outcome \((3, 1)\). Notice that this is not the same set as \(\{(1, 3)\}\). (But the set \(\{3, 1\}\) is the same as the set \(\{1, 3\}\).)
- The sample space of \(\{1, 2, 3, 4, 5, 6\}\) could correspond to a
*single*roll of a fair six-sided die. In this case, the event \(\{3, 1\}\) would be the event that the roll is either a 3 or a 1. (The set \(\{3, 1\}\) is the same as the set \(\{1, 3\}\).)

For the purposes of this text, \(\mathcal{F}\) can be considered to be the set of all subsets of \(\Omega\). Technically, \(\mathcal{F}\) is a

*\(\sigma\)-field*of subsets of \(\Omega\): \(\mathcal{F}\) contains \(\Omega\) and is closed under countably many elementary set operations (complements, unions, intersections). This requirement ensures that if \(A\) and \(B\) are “events of interest”, then so are \(A\cup B\), \(A\cap B\), and \(A^c\). While this level of technical detail is not needed, we prefer to introduce the idea of a “collection of events” now since a probability measure is a function whose input is an event (set) and not an outcome.↩There are 4 cards that could potentially go in box 1, then 3 cards that could potentially go in box 2, 2 to box 3, and 1 left for box 4. This results in \(4\times3\times2\times1=4! = 24\) possible outcomes. We will see more counting rules in Chapter

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