1.2 Events
1 (a) The probability of $P(b)$ $=$ $1$ $-$ $P(a)$ $-$ $P(c)$ $-$ $P(d)$ $-$ $P(e)$ $=$ $1$ $-$ $0.13$ $-$ $0.48$ $-$ $0.02$ $-$ $0.22$, and (b) the probaiblity of event $P(A)$ $=$ $p(c)$ $+$ $p(d)$ $=$ $0.48$ $+$ $0.02$ $=$ $0.5$. Now that $P(A)$ is known and the equation for computing the complement of $A$ is given in the book, then it can be said that (c) $P(A^\prime)$ $=$ $1 - P(A)$ $=$ $1 - 0.5$ $=$ $0.5$.
2 (a) If $P(A) = 0.27$, then $P(b)$ $=$ $P(A)$ $-$ $p(c)$ $-$ $p(e)$ $=$ $0.27$ $-$ $0.11$ $-$ $0.06$ $=$ $0.22$. Now that $P(b)$ is know, it becomes possible to see that (b) $P(A^\prime)$ $=$ $1 - P(A)$ $=$ $1$ $-$ $(P(b)$ $+$ $P(c)$ $+$ $P(e))$ $=$ $1$ $-$ $(0.22$ $+$ $0.11$ $+$ $0.06)$ $=$ $1$ $-$ $0.39$ $=$ $0.61$. Therefore (c) $P(d)$ $=$ $1$ $-$ $p(a)$ $-$ $p(b)$ $-$ $p(c)$ $-$ $p(e)$ $-$ $p(f)$ $=$ $1$ $-$ $0.09$ $-$ $0.22$ $-$ $0.11$ $-$ $0.06$ $-$ $0.29$ $=$ $0.23$.
3 If birthdays are equally likely to fall on any day, then the probability of one birthday to fall on one day out of the possible 365, excluding the leap years, is $\frac{1}{365}$. The probability of birthday falling to a specific month is $\prod_{n \ = \ 1}^{d}\frac{1}{365}$ $=$ $\frac{31}{365}$, where $d$ is the number of days in a month. For January $d$ $=$ $31$, and for February $d$ $=$ $28$, again, excluding the leap years, so the probability of having a birthday in January is $\frac{31}{365}$ and in February it’s $\frac{28}{365}$.
4 The costs will decrease when they are not staying the same or increasing. Since the given probabilities for the latter two are $0.018$ and $0.03$, subtracting them from the total probability $1$ is the probability of costs decreasing, i.e. $0.79$. Costs won’t increase when they stay the same or are decreaseng is $0.03$ $+$ $0.18$ $=$ $0.21$.
5 The situation can be modeled in the following way: Let the proposition $A$ be equal to that the Company A’s stock will increase, and proposition $B$ mean that the Company B’s stock will increase. Then there are four probabilities to consider which are given in the following Table 1,
$$ \begin{array}{ c c c } & & p \\[0.5em] \hline \quad A \uparrow & B \uparrow & 0.38 \quad \\[0.5em] \quad A \uparrow & B \downarrow & x \quad \\[0.5em] \quad A \downarrow & B \uparrow & 0.16 \quad \\[0.5em] \quad A \downarrow & B \downarrow & 0.11 \quad \\[0.5em] \end{array} $$where $X\uparrow$ means that $X$’s stock will rise, and $X\downarrow$ the opposite. The $p$ indicates the probability of events in a give row happening simultaneously.
Since the probabilities of all of these events must add to $1$, the
$$ x = 1 - 0.38 - 0.16 - 0.11 = 0.35, $$that is the event $A\uparrow$ and $B\downarrow$ has the probability of $0.35$.
The probability that at least one company’s stock will rise is the probability is the cumulative probability of those probabilities where it is true that at least one company is in state $\uparrow$. Adding those probabilities are
6 Writing out explicitly all the combinations of outcomes when rolling two fair six sided dices are given in the Table 2:
$$ \small { \begin{array}{ c c c c c c } (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\[0.5em] \textbf{(2,1)} & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\[0.5em] \textbf{(3,1)} & \textbf{(3,2)} & (3,3) & (3,4) & (3,5) & (3,6) \\[0.5em] \textbf{(4,1)} & \textbf{(4,2)} & \textbf{(4,3)} & (4,4) & (4,5) & (4,6) \\[0.5em] \textbf{(5,1)} & \textbf{(5,2)} & \textbf{(5,3)} & \textbf{(5,4)} & (5,5) & (5,6) \\[0.5em] \textbf{(6,1)} & \textbf{(6,2)} & \textbf{(6,3)} & \textbf{(6,4)} & \textbf{(6,5)} & (6,6) \\[0.5em] \end{array} } $$In the Table 2 all outcomes where all the strictly greater values of the red dice, i.e. the first value of a tuple $(x,y)$, are boldened. The number of these outcomes is $15$ out of all $36$ possible values, i.e. the probability of outcome where the red die is strictly less $0.5$ is
$$ \dfrac{15}{16} = 0.4167. $$Complement of the event is $1 - 0.4167 = 0.5833$.
7 In a deck of cards there are commonly $52$ cards. There are two possible colors in these kind of decks, so each color has $1 / 2$ $=$ $0.5$ proability being one of the either color.
8 From a deck of $52$ cards there are four possible aces, so the probability of card being a ace of one suit is $\frac{4}{52}$ $=$ $0.076923$.
9 (a) The probability that Terica is the winner is $\frac{1}{4} = 0.25$ and (b) the probability that Terica is winner or a runner up is $\frac{1}{4}$ $+$ $\frac{1}{4}$ $=$ $\frac{2}{4}$ $=$ $\frac{1}{2}$ $=$ $0.5$.
10 (a) The probability that the battery $\text{I}$ lasts longest is the sum of all of the outcomes where $\text{I}$ is at the third index. These are the outcomes at center right and bottom right of the Figure 1.24, so summing their probabilities are $0.39$ $+$ $0.03$ $=$ $0.42$.
(b) Similarily the outcomes where the battery $\text{I}$ is at the first index of the tuple are the probabilites that the battery $\text{I}$ last shortest, that is, the probabilities of top left and top right. Summing these probabilities are $0.11$ $+$ $0.07$ $=$ $0.18$, which is the the probability of the event that battery $\text{I}$ lasts the shortest.
(c) The event that battery $\text{I}$ does not last longest is the complement of the probability of battery $\text{I}$ lasting the longest, i.e.
$$ 1 - 0.42 = 0.58. $$(d) The probabilities of the event where the battery $\text{I}$ does lasts longer than battery $\text{II}$ are the sum of probabilities where $\text{I}$ lies on the right hand-side of $\text{II}$. These are center left and right, and the one on the bottom right, so the sum is $0.24$ $+$ $0.39$ $+$ $0.03$ $=$ $0.66$.
11 (a) The probability that both assembly lines are shut down $(S)$ are $p(S,S)$ $=$ $0.02$. (b) When neither of the assembly lines is down, a assembly line is running in full capacity ($F$) or in partial capacity $(P)$. These are the events where the value of the tuples in Figure 1.25 are either $F$ or $P$, so the sum $(P,P)$ $+$ $(P,F)$ $+$ $(F,P)$ $+$ $(F,F)$ $=$ $0.14$ $+$ $0.20$ $+$ $0.21$ $+$ $0.21$ $+$ $0.19$ $=$ $0.74$.
12 When a fair coin is thrown three times, the number of possible combinations of having heads ($\text{H}$) or tails ($\text{T}$) are
$$ \begin{array}{ c c c } \text{1} & \text{2} & \text{3} \\[0.5em] \hline \text{\textbf{H}} & \text{\textbf{H}} & \text{\textbf{H}} \\ \text{\textbf{H}} & \text{\textbf{H}} & \text{\textbf{T}} \\ \text{H} & \text{T} & \text{H} \\ \text{H} & \text{T} & \text{T} \\ \text{T} & \text{\textbf{H}} & \text{\textbf{H}} \\ \text{T} & \text{H} & \text{T} \\ \text{T} & \text{T} & \text{H} \\ \text{T} & \text{T} & \text{T} \\ \end{array} $$where all the toss-sequences that have two consequantial $\text{H}$ are boldened. There are three such rows. Also, as it can be seen from the table, in total there are eight possible combinations one can have from tossing a coin three times, so the probability of having two consecutive $\text{H}$ is
$$ \frac{3}{8} = 0.3750. $$13 The companys expectation that revenue is not below expectation is with probability $0.26$ $+$ $0.36$ $+$ $0.11$ $=$ $0.73$.
14 The probability that the advertising campaign is launched is the sum of the all probabilities that are above “canceled before launch”, i.e. the probability that the advertising campaign is launced but canceled ($0.18$), launched and runs its targeted length ($0.43$), and launched and is extended beyond its targeted length ($0.29$), which is $0.9$.