1.3 Combination of Events

1 Speaking in terms of sets, because $A = \{ 0 \}$ is an event of $\mathcal{S}$, it makes the (event) set $A$ to contain a single element, namely the element “$0$”. This means that the event $A$ has an element that indicates nothing and the element is itself something, so the set can not therefore be empty, so $A \neq \emptyset$.

2 (a) $P(B)$ $=$ $0.01$ $+$ $0.02$ $+$ $0.05$ $+$ $0.11$ $+$ $0.06$ $+$ $0.13$ $+$ $0.08$ $=$ $0.46$, (b) $P(B \cap C)$ $=$ $0.02$ $+$ $0.05$ $+$ $0.11$ $=$ $0.08$ (c) $P(A \cup C)$ $=$ $0.07$ $+$ $0.05$ $+$ $0.01$ $+$ $0.08$ $+$ $0.02$ $+$ $0.04$ $+$ $0.05$ $+$ $0.07$ $+$ $0.11$ $=$ $0.5$, (d) $P(A \cap B \cap C)$ $=$ $0.02$ $+$ $0.05$ $+$ $0.11$ $=$ $0.18$, (e) $P(A \cup B \cup C)$ $=$ $0.07$ $+$ $0.05$ $+$ $0.01$ $+$ $0.08$ $+$ $0.02$ $+$ $0.04$ $+$ $0.05$ $+$ $0.07$ $+$ $0.06$ $+$ $0.11$ $+$ $0.08$ $+$ $0.011$ $+$ $0.13$ $=$ $0.791$, (f) $P(A^\prime \cap B)$ $=$ $0.04$ $+$ $0.05$ $+$ $0.03$ $0.01$ $+$ $0.02$ $+$ $0.05$ $+$ $0.08$ $+$ $0.06$ $+$ $0.11$ $+$ $0.13$ $=$ $0.58$, (g) $P(B^\prime \cup C)$ $=$ $0.08$ $+$ $0.04$ $+$ $0.07$ $+$ $0.11$ $=$ $0.3$, (h) $P(A \cup (B \cap C))$ $=$ $0.07$ $+$ $0.05$ $+$ $0.01$ $+$ $0.08$ $+$ $0.02$ $+$ $0.05$ $+$ $0.11$ $=$ $0.39$, (i) $P((A \cup B) \cap C)$ $=$ $0.07$ $+$ $0.05$ $+$ $0.01$ $+$ $0.06$ $+$ $0.08$ $+$ $0.13$ $=$ $0.4$, and (j) $P(A^\prime \cup C)^\prime$ $=$ $0.04$ $+$ $0.05$ $+$ $0.07$ $+$ $0.11$ $+$ $0.11$ $+$ $0.03$ $=$ $0.38$

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4 Given that the event $A$ means a person is a female, $B$ that a person has black hair and $C$ that a person has blue eyes, event (a) $A \cap B$ means that person is a black haired female, (b) $A \cup C^\prime$ means that person is female, has brown eyes or both, (c) $A^\prime \cap B \cap C$ means that person is non-female with black hair and brown eyes, and (d) $A \cap (B \cup C)$ means that person is female with either black hair, blue eyes or both.

5 Let the event $A$ be that the card drawn from a deck of cards is either $\diamondsuit$ or $\heartsuit$, i.e., a red suit, and the event $B$ be that the card is either $\clubsuit$ or $\spadesuit$, i.e., a black suit. Then if a card is drawn from a deck of cards, the card is either from $A$ or from $B$, but the card can not be from both sets at the same time, so $A \cap B = \emptyset$, making the events mutually exclusive by definition (given in the book at page $17$). On the other hand, if the drawn card is ace, say, for example that $A$ $=$ $\{$ $\mathcal{A}\clubsuit,$ $\mathcal{A}\diamondsuit,$ $\mathcal{A}\heartsuit,$ $\mathcal{A}\spadesuit$ $\}$ and that the drawn card is $B$ $=$ $\{$ $\mathcal{A}\heartsuit,$ $2\heartsuit,$ $\ldots,$ $\mathcal{K}\heartsuit$ $\}$, then $A \cap B$ $=$ $\{ \ \mathcal{A}\heartsuit \ \}$ $\neq$ $\emptyset$, so these events are not mutually exclusive.

6 The events are not mutually exclusive since the intersection $A \cap B$ has a probability value or the intersection is not an empty set. Assuming that $A^\prime \cup B^\prime$ $=$ $\emptyset$,

$$ \begin{array}{ r c l } 1 & = & P(A) + P(B) - P(A \cap B) \\[0.5em] & = & 0.4 + P(B) - 0.3 \end{array} $$

$$ \phantom{.} \ P(B) = 1 - 0.4 - 0.3 = 0.4 \ , $$

which is a possible value for probability $P(B)$.

7 Because

$$ P(A \cup B) = P(A) + P(B) - P(A \cap B) $$

then, given the probabilities of the exercise,

$$ \phantom{,} \ 0.8 = 0.5 + P(B) - 0.1 \ , $$

so solving for $P(B)$ it can be shown that

$$ \phantom{.} \ P(B) = 0.8 - 0.5 + 0.1 = 0.4 \ . $$

Substituting the probability $P(B)$ back to origianal equation it can be seen that $0.8$ $=$ $0.5$ $+$ $0.4$ $-$ $0.1$ $=$ $0.8$, that is, the probability value of $P(B)$ solves the problem, so $P(B) = 0.4$.

8 Let the event that evaluation reveals a problem be named as $A$ and the event that evaluation is accurate, i.e. correct, as $B$. The given probabilities can then be written as $P(B)$ $=$ $0.85$, $P(A \cap B^\prime)$ $=$ $0.10$ and $P(A^\prime \cap B)$ $=$ $0.25$ .

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9 In order for two events to be mutually exclusive, the intersection of the events is an empty set. In other words, given two events $A$ and $B$, $A \cap B$ $=$ $\emptyset$. Two or more mutually exclusive events do not have common outcomes. When the events $A$, $B$ ja $C$ are defined as in the exercise, the task is to find the intersection $A \cap B \cap C$. Since a card drawn from a deck of cards can only be of one suit, these events do not have any common outcomes so their intersection is empty making the events mutually exclusive.

The probability of card drawn from a deck of cards is one of the four suits is $\frac{1}{4}$, so

$$ \phantom{.} \ P(A \cup B \cup C) = P(A) + P(B) + P(C) = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4} \ . $$

$$ \begin{array}{c c c} & P(A \cup B \cup C) & \phantom{=} \\[0.75em] = & P(A) + P(B) + P(C) \\[0.5em] = & \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} \\[0.5em] = & \phantom{.} \ \dfrac{3}{4} \ . \end{array} $$

The event $A^\prime$ defines the outcomes of drawing any other card than a card from the suit $\heartsuit$. Since the $B$ is in this set, i.e. $B$ is a outcome in $A^\prime$, because $B$ means the event of obtaining a $\clubsuit$, $B \subset A^\prime$.

10 The outcomes of (a) $A$ $\cap$ $B$ is the event where a card drawn from a deck of cards is $\mathcal{A}\heartsuit$ or $\mathcal{A}\diamondsuit$, (b) the event $A$ $\cup$ $B$ is the event that the card is ace or a picture card, (c) the event $B$ $\cap$ $C^\prime$ the event that card is red suited number card and (d) the event $A \cup (B^\prime \cap C)$ that the card is black pictured card and or ace or not. Since a card cannot be picture card and a ace at the same time, the event could be taken to mean event where drawn card is a picture card of black suit.

11 Let $A$ mean that repair is on time, $B$ that repair is satisfactory and $A \cap B$ that repair is on time and satisfatory. Then $A^\prime \cap B^\prime$ means that repair wasn’t on time and was unsatisfactory, that is, the event is the intersection of situations when repair is on time and not satisfactory. The probability of this event is

$$ \begin{array}{r c l} P(A^\prime \cap B^\prime) & = & 1 - P(A \cup B) \\[0.5em] & = & 1 - (P(A) - P(B) + P(A \cap B)) \\[0.5em] & = & 1 - 0.74 - 0.41 + 0.26 \\[0.5em] & = & 0.11. \end{array} $$

$$ \begin{array}{c l} & \hspace{-2.0em} P(A^\prime \cap B^\prime) \\[0.5em] = & 1 - P(A \cup B) \\[0.5em] = & 1 - P(A) - P(B) + P(A \cap B) \\[0.5em] = & 1 - 0.74 - 0.41 + 0.26 \\[0.5em] = & 0.11. \end{array} $$

12 The probability of the event $A$, that is, ball is shiny is $P(A)$ $=$ $\frac{91}{200}$ $=$ $0.455$, the probability of the event $B$, that is, ball is red is $P(B)$ $=$ $\frac{79}{200}$ $=$ $0.395$ and the probability $A \cap B$, that is, ball is shiny and red is $P(A \cap B)$ $=$ $\frac{55}{200}$ $=$ $0.275$.

The probability of event $(A \cup B)$, that is, a randomly selected ball is either a shiny ball or a red ball is

$$ \begin{array}{r c l} P(A \cup B) & = & P(A) + P(B) - P(A \cap B) \\[0.5em] & = & 0.455 + 0.395 - 0.275 \\[0.5em] & = & 0.575. \end{array} $$

$$ \begin{array}{c l} & \hspace{-2em} P(A \cup B) \\[0.5em] = & P(A) + P(B) - P(A \cap B) \\[0.5em] = & 0.455 + 0.395 - 0.275 \\[0.5em] = & 0.575. \end{array} $$

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14 The probabilites events can be declared as $P(E_1) = 0.08$ revenue is below expectation, $P(E_2) = 0.19$ revenue is slightly below expectation, $P(E_3) = 0.26$ revenue meets the expectation, $P(E_5) = 0.36$ revenue is slightly above expectation, and $P(E_6) = 0.11$ revenue is above expectation.

Given the probabilities, the union of them is

$$ \begin{array}{c l} & P \left( \displaystyle \bigcup_{i \ = \ 1}^6 E_i \right) \\[0.5em] = & P(E_1 \cup E_2 \cup E_3 \cup E_4 \cup E_5 \cup E_6) \\[0.5em] = & P(E_1) \cup P(E_2) \cup P(E_3) \cup P(E_4) \cup P(E_5) \cup P(E_6) \\[0.5em] = & P(E_1) + P(E_2) + P(E_3) + P(E_4) + P(E_5) + P(E_6) \\[0.5em] = & 0.08 + 0.19 + 0.26 + 0.36 + 0.11 \\[0.5em] = & 1, \end{array} $$

$$ \begin{array}{c l} & P \left( \displaystyle \bigcup_{i \ = \ 1}^6 E_i \right) \\[0.5em] = & P(E_1 \cup E_2 \cup E_3 \cup E_4 \cup E_5 \cup E_6) \\[0.5em] = & P(E_1) \cup P(E_2) \cup P(E_3) \\[0.5em] & \cup P(E_4) \cup P(E_5) \cup P(E_6) \\[0.5em] = & P(E_1) + P(E_2) + P(E_3) \\[0.5em] & + P(E_4) + P(E_5) + P(E_6) \\[0.5em] = & 0.08 + 0.19 + 0.26 + 0.36 + 0.11 \\[0.5em] = & 1, \end{array} $$

so at the intersection of these events is no outcomes, or in other words, the intersection of these events is empty, so statistically speaking, the probability of any combination of these events happening at the same time is zero.

The union of the given events $A$ and $B$, or in other words, $A \cup B$ is the event that the event is not below and not above expectation, that is, $A \cup B$ is the event that revenue meets expectation. The probability of this event is $0.26$.

14 In the context of the exercise, and usually in general, the possible revenue outcomes are mutually exclusive. This means that in the intersection of events $A$ (revenue is not below expection, i.e. revenue meets expectation or is above expectation) and $B$ (revenue is not above expectation, i.e. the revenue meets expectation or is below) is that event meets expectation. The union of these events mean all possible outcomes: Revenue is considerably or slightly below expectation, meets the expectation, or is slightly or considerably above expectation.

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