2.2 Events

An event is something that might happen. For example, if we’re interested in the weather conditions in our city tomorrow, events include

  • the high temperature is 75°F
  • the high temperature is above 75°F
  • it rains
  • it does not rain
  • it rains and the high temperature is above 75°F

An outcome consists of all the information about tomorrow’s weather conditions, while an event is a collection of outcomes that satisfy some criteria.

Definition 2.2 An event \(A\) is a subset of the sample space: \(A\subseteq \Omega\). If the random phenomenon yields outcome \(\omega\), we say “event \(A\) occurred” if \(\omega\in A\). The collection of all events of interest26 is denoted \(\mathcal{F}\).

The sample space is the collection of all possible outcomes. An event represents only those outcomes which satisfy some criteria. 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.

Example 2.9 Roll a four-sided die twice, and record the result of each roll in sequence. Using the sample space from Example 2.1, identify the following events.

  1. \(A\), the event that the sum of the two dice is 4.
  2. \(B\), the event that the sum of the two dice is at most 3.
  3. Donny Don’t says “we should just consider the sample space to be \(\{2, \ldots, 8\}\)” so that \(A = 4\) and \(B = \{2, 3\}\). Do you agree? (Also, can you spot the subtle mistake that Donny made?)
  4. Identify \(C\), the event that the larger of the two rolls (or the common roll if a tie) is 3
  5. Identify and interpret \(A\cap C\).
  6. Donny Don’t says that \(D\), the event that the first roll is a 3, is \(D=\{3\}\). Explain to Donny his mistake, and identify event \(D\).
  7. Identify \(E\), the event that the second roll is a 3.
  8. Identify and interpret \(D \cap E\).
  9. Identify and interpret \(D \cup E\).
  10. If the outcome is \((1, 3)\), which of the events above occurred?

Solution. to Example 2.9

Show/hide solution
  1. Remember that the sample space consists of 16 possible ordered pairs of rolls \[\begin{align*} \Omega & = \{(1, 1), (1, 2), (1, 3), (1, 4),\\ & \qquad (2, 1), (2, 2), (2, 3), (2, 4),\\ & \qquad (3, 1), (3, 2), (3, 3), (3, 4),\\ & \qquad (4, 1), (4, 2), (4, 3), (4, 4)\} \end{align*}\] (See also Table 2.1.) All events must be defined as subsets of this sample space. So \(A=\{(1, 3), (2, 2), (3, 1)\}\) is the event that the sum of the two dice is 4.
  2. \(B=\{(1, 1), (1, 2), (2, 1)\}\) is the event that the sum of the two dice is at most 3.
  3. Tell Donny it’s better to use the sample space from Example 2.1 since 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 events like whether the larger of the two rolls is a 3. (Donny has also made a subtle mistake in writing \(A = 4\). An event is always a set, even if it contains just a single outcome. Given Donny’s sample space, he should have written \(A = \{4\}\).)
  4. \(C=\{(1, 3), (2, 3), (3, 1), (3, 2), (3, 3)\}\), the event that the larger of the two rolls (or the common roll if a tie) is 3
  5. \(A\cap C=\{(1, 3), (3, 1)\}\) is the event that both the sum of the two dice is 4 and the larger of the two rolls is 3.
  6. Tell Donny 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. (Remember, there is always a single sample space upon which all events are defined.) \(D=\{(3, 1), (3, 2), (3, 3), (3, 4)\}\) is the event that the first roll is a 3.
  7. \(E=\{(1, 3), (2, 3), (3, 3), (4, 3)\}\), the event that the second roll is a 3. Note that this is not the same event as \(D\).
  8. \(D \cap E = \{(3, 3)\}\) is the event that both rolls result in a 3. That is, (3, 3) is the only outcome that satisfies both events \(D\) and \(E\). While an event is always a set, it can be a set consisting of a single outcome (or the empty set).
  9. \(D \cup E = \{(3, 1), (3, 2), (3, 3), (3, 4), (1, 3), (2, 3), (4, 3)\}\) is the event that at least one of the two rolls results in a 3. Notice that the union is inclusive: \((3, 3)\), the outcome that satisfies both \(D\) and \(E\), is an element of \(D\cup E\). But also notice that the outcome \((3, 3)\) only appears once in the set \(D \cup E\).
  10. 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.

We reiterate (again!) that there is a single sample space, upon which all events are defined. In the above example, events that involved only the first or second roll such as \(D\) and \(E\) were still defined in terms of pairs of rolls. 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.

While an event is always a set, it can be a set consisting of a single outcome, or a set consisting of no outcomes at all (the empty set \(\emptyset\)).

Definition 2.3 Two events \(A\) and \(B\) are disjoint (a.k.a. mutually exclusive) if \(A\cap B=\emptyset\). That is, \(A\) and \(B\) are disjoint if they have no outcomes in common.

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 \(i \neq j\). That is, multiple events are disjoint if none of the events have any outcomes in common.

Example 2.10 Continuing Example 2.9.

  1. Identify two events that are disjoint.
  2. Are the events \(A, B, C\) disjoint?

Solution. to Example 2.10

Show/hide solution
  1. Events \(A\) and \(B\) are disjoint. It is not possible for an outcome to satisfy both criteria “the sum is 4” and “the sum is at most 3”.
  2. No, because \(A\cap C \neq \emptyset\) since there are outcomes that satisfy both criteria “the sum is 4” and “the larger roll is 3”.

Example 2.11 Consider the outcome of a sequence of 4 flips of a coin. Using the sample space from Example 2.2, identify the following events.

  1. Identify \(A\), the event that exactly 3 of the flips land on heads.
  2. Identify \(B\), the event that exactly 4 of the flips land on heads.
  3. Identify \(C\), the event that the at least 3 of the flips land on heads. How does \(C\) relate to \(A\) and \(B\)?
  4. 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 \(C = \{3, 4\}\).
  5. Identify \(D\), the event that at least 3 heads are flipped in a row.
  6. Identify \(E\), the event that the first two flips results in tails.
  7. Identify \(D\cap E\).
  8. If the outcome is HHTH, which of the events above occurred?

Solution. to Example 2.11

Show/hide solution
  1. \(A = \{HHHT, HHTH, HTHH, THHH\}\) is the event that exactly 3 of the flips land on heads
  2. \(B = \{HHHH\}\), 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.
  3. \(C = \{HHHT, HHTH, HTHH, THHH, HHHH\}\) is the event that the at least 3 of the flips land on heads. Also \(C = A \cup B\).
  4. 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.
  5. \(D = \{HHHT, THHH, HHHH\}\) is the event that at least 3 heads are flipped in a row. Note that \(D\) is not the same event as \(C\).
  6. \(E = \{TTHH, TTHT, TTTH, TTTT\}\) 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.
  7. \(D\cap E=\emptyset\) 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.
  8. If the outcome is HHTH then events \(A\) and \(C\) occur. Events \(B\), \(D\), \(E\), and \(D \cap E\) do not occur.

Example 2.12 (Matching problem) Rocks labeled 1, 2, 3, 4, are placed at random in spots labeled 1, 2, 3, 4, with spot 1 the correct spot for rock 1, etc. Using the sample space from Example 2.3, identify the following events.

  1. \(A\), the event that all rocks are put in the correct spot.
  2. \(B\), the event that no rocks are put in the correct spot.
  3. \(C\), the event that exactly 3 rocks are put in the correct spot.
  4. \(D\), the event that rock 3 is put (correctly) in spot 3.

Solution. to Example 2.12

Show/hide solution
  1. Recall that each outcome is a particular placement of rocks in the spots. For example, the outcome 3214 (or \((3, 2, 1, 4)\)) represents that rock 3 is put in spot 1, rock 2 in spot 2, rock 1 in spot 3, and rock 4 in spot 4. There is only one outcome, 1234, for which all rocks are put in the correct spot, so \(A=\{1234\}\). Remember that an event is always a set, but it can be a set consisting of a single outcome.
  2. For each outcome in the sample space check to see if the criteria holds to identify \(B=\{2143, 2341, 2413, 3142, 3412, 3421, 4123, 4312, 4321\}\) as the event that no rocks are put in the correct spot.
  3. There are no outcomes in which exactly 3 rocks are put in the correct spot so \(C=\emptyset\). (If three rocks are in their correct spots, then the remaining rock must be in its correct spot too.)
  4. \(D=\{1234, 1432, 2134, 2431, 4132, 4231\}\) is the event that rock 3 is put (correctly) in spot 3. Even though event \(D\) only concerns rock 3, since the sample space consists of the placements of each of the rocks then all events must be expressed in terms of these outcomes.

When more than just a few events are of interest, subscripts are commonly used to identify different events, as in the following example.

Example 2.13 Recall Example 2.5 in which 3 phone numbers are selected from a bank that contains 5 phone numbers, labeled {1, 2, 3, 4, 5}. Let \(A_i\) be the event that phone number \(i\) is selected for the sample, \(i=1, 2, 3, 4, 5\). For example, \(A_4\) is the event that phone number 4 is selected for the sample. Using the sample space from Example 2.5 identify the following events, both in terms of \(A_i\)’s and as subsets of the sample space.

  1. The event that phone number 1 is selected for the sample.
  2. The event that phone number 4 is selected for the sample.
  3. The event that phone numbers 1 and 4 are selected for the sample.
  4. The event that phone numbers 1 or 4 are selected for the sample.
  5. The event that phone numbers 1, 4, and 5 are selected for the sample.
  6. The event that phone numbers 1, 4, or 5 are selected for the sample.

Solution. to Example 2.13

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  1. The event \(A_1=\{\{1, 2, 3\}, \{1, 2, 4\}, \{1, 2, 5\}, \{1, 3, 4\}, \{1, 3, 5\}, \{1, 4, 5\}\}\) consists of the samples which include phone number 1. Even though event \(A_1\) only concerns phone number 1, since the sample space consists of sets (samples) of size 3 then all events must be expressed in terms of these outcomes.
  2. The event \(A_4=\{\{1, 2, 4\}, \{1, 3, 4\}, \{1, 4, 5\}, \{2, 3, 4\}, \{2, 4, 5\}, \{3, 4, 5\}\}\) consists of the samples which include phone number 4.
  3. The event \(A_1\cap A_4=\{\{1, 2, 4\}, \{1, 3, 4\}, \{1, 4, 5\}\}\) consists of the samples which include both phone numbers 1 and 4.
  4. The event \(A_1\cup A_4=\{\{1, 2, 3\}, \{1, 2, 4\}, \{1, 2, 5\}, \{1, 3, 4\}, \{1, 3, 5\}, \{1, 4, 5\}, \{2, 3, 4\}, \{2, 4, 5\}, \{3, 4, 5\}\}\) consists of the samples which include either phone number 1 or 4 (or both).
  5. The event \(A_1\cap A_4\cap A_5=\{\{1, 4, 5\}\}\) consists of the only outcome in which phone numbers 1, 4, and 5 are selected.
  6. The event \(A_1\cup A_4\cup A_5=\Omega\) is the event that phone numbers 1, 4, or 5 are selected for the sample. Since only two of the five numbers are not selected from the sample, at least one of the numbers 1, 4, or 5 must be included in the sample.

Remember that intervals of real numbers such as \((a,b), [a,b], (a,b]\) are also sets, and so can also be events. For example, if an outcome is the result of a single spin of the spinner in Figure 2.2, events include

  • \([0, 0.5]\), the result is between 0 and 0.5 (the needle lands in the right half of the spinner)
  • \([0.75, 1]\), the result is between 0.75 and 1 (the needle lands in the northwest quarter of the spinner)
  • \([0.595, 0.605)\), the result rounded to two decimal places is 0.6
  • \(\{0.6\}\), the result is 0.6 exactly (the needle points exactly at 0.60000000\(\ldots\))

It is often helpful to conceptualize and visualize events (sets) with pictures, as in the following example.

Example 2.14 (Meeting problem) Regina and Cady plan to meet for lunch between noon and 1 but they are not sure of their arrival times. Using the sample space from Example 2.6, identify the following events using both pictures and mathematical notation. (Hint: see Figure 2.1.)

  1. Identify \(A\), the event that Regina arrives after Cady.
  2. Identify \(B\), the event that either Regina or Cady arrives before 12:30.
  3. Identify \(C\), the event that Cady arrives first and Regina arrives at most 15 minutes after Cady.
  4. Identify \(D\), the event that Regina arrives before 12:24.

Solution. to Example 2.14

Show/hide solution
  1. See Figure 2.4 for pictures. Recall that an outcome is a pair of values \(\omega = (\omega_1, \omega_2)\) corresponding to the arrival times of (Regina, Cady), and that \([0, 1]\) corresponds to the one hour time interval from noon (0) to 1. \(A = \{(\omega_1, \omega_2): \omega_1>\omega_2\}\) is the event that Regina arrives after Cady. (Throughout 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.)
  2. \(B = \{(\omega_1, \omega_2): \omega_1<0.5 \text{ or } \omega_2<0.5\}\) is the event that either Regina or Cady arrives before 12:30. This event is the complement of the event that both arrive after 12:30, \(B = ([0.5, 1]\times[0.5, 1])^c\). The event \(B\) can also be written as \(\{(\omega_1, \omega_2): \min(\omega_1,\omega_2)<0.5\}\), the event that the earlier of the two arrival times is before 12:30.
  3. \(C = \{(\omega_1, \omega_2): \omega_2<\omega_1\le \omega_2+0.25\} = \{(\omega_1, \omega_2): \omega_1 > \omega_2\ge \omega_1 - 0.25\}\) is the event that Regina arrives at most 15 minutes after Cady (and Cady arrives first).
  4. \(D = \{(\omega_1, \omega_2): \omega_1<0.4\}\) is the event that Regina arrives before 12:24. Even though event \(D\) only concerns Regina, since the sample space consists of pairs of arrival times then all events must be expressed in terms of these outcomes.
Illustration of the events in Exercise 2.14. The square represents the sample space \(\Omega=[0,1]\times[0,1]\).

Figure 2.4: Illustration of the events in Exercise 2.14. The square represents the sample space \(\Omega=[0,1]\times[0,1]\).

Example 2.15 (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 understand the correct answers.

  1. The possible rolls are 1 through 6, so the sample space is \(\{1, 2, 3, 4, 5, 6\}\).
  2. 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 \(\{10, 11, 12\}\).
  3. The event that the first roll is a 3 is \(\{3\}\).
  4. The event that the first roll is a 3 and the second roll is a 1 is \(\{3, 1\}\)
  5. 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 2.15

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  1. 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. \[\begin{align*} \Omega = & \{ (1, 1), (1, 2), \ldots, (1, 6),\\ & \;\; (2, 1), (2, 2), \ldots, (2, 6),\\ & \;\; \vdots\qquad \qquad \quad \cdots \qquad \vdots\\ & \;\; (6, 1), (6, 2), \ldots, (6, 6) \} \end{align*}\]
  2. 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.
  3. 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.
  4. 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 \(\{1, 3\}\).)
  5. 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 single roll results in either either a 3 or a 1. (The set \(\{3, 1\}\) is the same as the set \(\{1, 3\}\).)

In many situations it is not possible to explicitly define a sample space in a compact way, and so outcomes and events are often only vaguely defined.
Nevertheless, there is always a sample space in the background representing possible outcomes, and collections of these outcomes represent events of interest.

Example 2.16 Recall Example 2.8. Customers enter a deli and take a number to mark their place in line. When the deli opens the counter starts 0; the first customer to arrive takes number 1, the second 2, etc. We record the counter over time, continuously, as it changes as customers arrive. Time is measured in minutes after the deli opens (time 0).

Using the sample space described in the solution for Example 2.8, describe in words how you might represent the following events. Sketch pictures like those in Figure 2.3 to represent the events.

  1. \(A\), the event that the first customer arrives after 2 minutes after the deli opens.
  2. \(B\), the event that at 5 minutes after opening, exactly 4 customers have arrived.

Solution. to Example 2.16

Show/hide solution
  1. Event \(A\) consists of paths like those in the plot on the left in Figure 2.5. The event consists of paths that start with a height of 0 and stay and a height of 0 after 2 minutes. In other words, event \(A\) consists of all paths where the first jump occurs after time 2. The plot displays only ten outcomes that satisfy event \(A\); the event consists of all such outcomes.
  2. Event \(B\) consists of paths like those in the plot on the right in Figure 2.5. The event consists of all paths that have a height of 4 at time 5. In other words, event \(B\) consists of all paths where the fourth jump (fourth customer to arrive) occurs before time 5 but the fifth jump (fifth customer to arrive) occurs after time 5. The plot displays only ten outcomes that satisfy event \(B\); the event consists of all such outcomes.
A few outcomes illustrating the events in Example 2.16. Left: Ten outcomes that satisfy the event that the first customer arrives after 2 minutes. Right: Ten outcomes that satisfy the event that at 5 minutes after opening, exactly 4 customers have arrived.A few outcomes illustrating the events in Example 2.16. Left: Ten outcomes that satisfy the event that the first customer arrives after 2 minutes. Right: Ten outcomes that satisfy the event that at 5 minutes after opening, exactly 4 customers have arrived.

Figure 2.5: A few outcomes illustrating the events in Example 2.16. Left: Ten outcomes that satisfy the event that the first customer arrives after 2 minutes. Right: Ten outcomes that satisfy the event that at 5 minutes after opening, exactly 4 customers have arrived.

2.2.1 The collection of events of interest

An event \(A\) is a set. The collection \(\mathcal{F}\) of “events of interest” is a collection of sets. For the purposes of this text, \(\mathcal{F}\) can be considered to be the set of all subsets of \(\Omega\).

As an example, consider a single roll of a four-sided die. (Don’t confuse this scenario with previous examples that involved two rolls.) The sample space consists of the four possible outcomes \(\Omega = \{1, 2, 3, 4\}\). (Again, don’t confuse this scenario with previous examples.) Any subset of this sample space is an event. The following table lists the 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 observing 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

A random phenomenon corresponds to a single sample space, but there are many events of interest. Listing the collection of all possible events as in the previous table is rarely done in practice, but we do so here to provide a concrete example of \(\mathcal{F}\).

We will see soon that we assign probabilities to events, rather than to outcomes. The collection of “events of interest” includes all the events that we will assign probabilities to.

2.2.2 Summary

  • An outcome \(\omega\) is a point.
  • The sample space \(\Omega\) is the set of all possible outcomes.
  • An event \(A\) is a collection of outcomes that satisfy some particular criteria. That is, an event is a subset of the sample space, \(A\subseteq\Omega\).
  • There are many events (sets) of interest associated with a random phenomenon. This collection of events (sets) is what \(\mathcal{F}\) represents.
  • All events of interest are defined in terms of a single sample space.
  • An event can be a set consisting of a single outcome, or no outcomes at all (the empty set \(\emptyset\)).
  • 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.
  • Pictures can be used to conceptualize and visualize events.

2.2.3 Exercises

  1. In Example 2.4, suppose we only buy 3 packages and we consider as our sample space outcome the results of just these 3 packages (prize in package 1, prize in package 2, prize in package 3). For example, 323 (or (3, 2, 3)) represents prize 3 in the first package, prize 2 in the second package, prize 3 in the third package. Then the sample space consists of 27 outcomes.

    \[ \Omega = \{111, 112, 113, 121, 122, 123, 131, 132, 133,\\ 211, 212, 213, 221, 222, 223, 231, 232, 233,\\ 311, 312, 313, 321, 322, 323, 331, 332, 333\} \]

    Identify the following events.

    1. \(A\), the event that we obtain a complete collection of prizes.
    2. \(B\), the event that we obtain only a single prize.
    3. \(C\), the event that we obtain only prize 1.
    4. \(D\), the event that two packages contain prize 1.
    5. \(E\), the event that no packages contain prize 1.

  2. Randomly select a baseball game and record the total number of runs scored and the length of the game. Identify the following events.

    1. \(A\), the event that exactly one run is scored and the game is less than 3 hours.
    2. \(B\), the event that exactly one run is scored.
    3. \(C\), the event that the game is less than 3 hours.

  3. Continuing Example 2.14, identify the following events with both pictures and mathematical notation.

    1. \(E_1\), the event that Regina and Cady arrive within 10 minutes of each other.
    2. \(E_2\), the event that the earlier to arrive has to wait more than 10 minutes for the other to arrive.
    3. \(E_3\), the event that at least one of Regina or Cady arrives after 12:40.
    4. \(E_4\), the event that Cady precisely at 12:34.

  4. In Example 2.7 consider the sample space \(\Omega = [200, 800] \times [200, 800]\) corresponding to possible pairs of (Math, Reading) scores. Identify the following events with both pictures and mathematical notation.

    1. \(A\), the event that the Math score is more than 100 points above the Reading score.
    2. \(B\), the event that the Math and Reading scores are equal.
    3. \(C\), the event that the sum of the Math and Reading scores is between 1200 and 1400.
    4. \(D\), the event that the sum of the Math and Reading scores is between 1200 and 1400 and the Math score is more than 100 points above the Reading score.
    5. \(E\), the event that the sum of the Math and Reading scores is between 1200 and 1400 and the Math score is above 700.

  5. Two players, A and B, play a single game of rock, paper, scissors (RPS). Record an outcome as \((a, b)\), where \(a\) represents player A’s throw and \(b\) player B’s, so the sample space consists of 9 outcomes \[ \Omega = \{(R, R), (R, P), (R, S), (P, R), (P, P), (P, S), (S, R), (S, P), (S, S)\} \]

    1. Specify the event that player A wins.
    2. Specify the event that player B wins.
    3. Specify the event that there is a tie, both directly and expressed in terms of the events from previous parts.
    4. Specify the event that player A throws rock.
    5. Specify the event that player A throws rock and does not lose, both directly and expressed in terms of the events from previous parts.

  1. 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) rather than an outcome (point).↩︎