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\begin{document}

\maketitle

\pagebreak

\begin{homeworkProblem}
A jar contains four coins: a nickel (5 cents), a dime (10 cents), a quarter (25 cents), and a half-dollar (50 cents).
Three coins are randomly selected without replacement from the jar.
\begin{enumerate}[label=(\alph*)]
    \item List all the possible outcomes in sample space S.
    \item What is the probabilty that the selection will contain the half-dollar?
    \item What is the probability that the total amount drawn will equal 65 cents or less?
    \item Which interpretations of probability are you applying?
\end{enumerate}

\solution

\part

$$S = \{(5,10,25), (5,25,50), (10, 25,50), (5,10,50)\}$$

\part

$$P(\text{half dollar}) = \frac{3}{4}$$

\part

Sum of the outcomes:
$$S = \{40, 80, 85, 65\}$$

$$P(\text{total} \leq 65) = \frac{2}{4} = \frac{1}{2}$$

\part

Classical interpretation.


\end{homeworkProblem}
\pagebreak

\begin{homeworkProblem}
    In a genetics experiment, the researcher mated two Drosophila fruit flies
    and observed the traits of 300 offsprint. The results are shown in the table

    \includegraphics[width=\textwidth]{q2.png}
    One of these offspring is randomly selected and observed for the two genetic traits

    \begin{enumerate}[label=(\alph*)]
    \item What is the probability that the fly has normal eye color and normal wing size?
    \item What is the probability that the fly has vermillion eyes?
    \item What is the probability that the fly has either vermillion eyes or miniature wings, or both?
     \item Which interpretations of probability are you applying?
\end{enumerate}

\solution

\part

$$P(\text{Normal Eye Color} \cap \text{Normal Wing Size}) = \frac{140}{300} = \frac{7}{15} = 0.46667$$

\part

$$P(\text{Vermillion Eyes}) = \frac{3+151}{300} = \frac{77}{150} = 0.513$$

\part

$$
\begin{align}
P(\text{vermillion eyes } \cup \text{ miniature wings})&= P(\text {vermillion eyes}) + P(\text{miniature wings}) - P(\text{vermillion eyes} \cap \text{miniature wings})\\
                                                       &= \frac{154}{300} + \frac{157}{300} - \frac{151}{300}
                                                       &= \frac{160}{300}
                                                       &= \frac{8}{15}
\end{align}
$$

\part

Relative frequency interpretation.

\end{homeworkProblem}
\pagebreak

\begin{homeworkProblem}
    City residents were surgeyed recently to determine readership of newspapers.
    50\% of the residents read the morning paper, 60\% read the evening paper and
    20\% read both newspapers. Find the probability that a resident selected
    at random reads the morning or evening paper

    \solution

    Given: \\
    $P(\text{morning}) = 0.5$\\
    $P(\text{evening}) = 0.6$\\
    $P(\text{both}) = 0.2 = P(\text{morning } \cap \text{ evening})$\\

    To find: $P(\text{morning or evening})$\\

    $$
    \begin{align}
        P(\text{morning} \cup \text{evening}) &= P(\text{morning}) + P(\text{evening}) - P(\text{morning} \cap \text{evening})$\\
                                              &= 0.5 + 0.6 - 0.2\\
                                              &= 0.9
    \end{align}
    $$

\end{homeworkProblem}
\pagebreak

\begin{homeworkProblem}
    The experience of a train passenger is that the train is cancelled with probability 0.02
    When it does run, it has probability of 0.9 of being on time.
    \begin{enumerate}[label=(\alph*)]
    \item Find the probability that the train runs on time
    \item Given that the train is not on time, find the probability that it has been cancelled
    \end{enumerate}

    \solution

    Given:\\
    P(\text{cancelled}) = 0.02\\
    P(\text{cancelled}^c) = 0.98\\
    P(\text{on time} | \text{cancelled}^c) = 0.9\\

    \part

    $$P(\text{on time}) =  0.98 \cdot 0.9 = 0.882$$

    \part

    From part (a):
    $$P(\text{on time}^c) = 1 - 0.882 = 0.118$$\\

    $$
    \begin{align}
        P(\text{cancelled} | \text{on time}^c) &= \frac{P(\text{cancelled} \cap \text{on time}^c)}{P(\text{on time}^c)}\\
            &= \frac{0.02\cdot 1}{0.02 \cdot 1 + 0.98 \cdot 0.1}\\
            &= 0.1694915\\
            &\approx 0.1694
    \end{align}
    $$

    Draw the probability tree diagram to visualise.\\
    Note: Once the train is cancelled, there is \textbf{NO WAY} the train can be on time,\\
    Therefore, $P(\text{on time} | \text{cancelled}) = 0$ and $P(\text{on time}^c | \text{cancelled}) = 1$.

\end{homeworkProblem}
\pagebreak

\begin{homeworkProblem}
    In a group of students 60\% are female and 40\% are male.
    A third of the female students study french but only a quarter of the male students study french
    \\
    A student is chosen at random from the group.
    \begin{enumerate}[label=(\alph*)]
    \item Show that the probability that the chosen student is female and studies french is 0.2
    \item Calculate the probability that the chosen student studies french
    \item Given that the chosen student does study french, calculate the conditional probability that the chosen student is female
    \end{enumerate}

    \solution

    Let f: female\\
    fr: studies french
    \\\\
    Given:
    $P(f) = 0.6$\\
    $P(f^c) = 0.4$\\
    $P(fr) = 0.3$\\
    $P(fr^c) = 0.7$\\
    $P(fr | f) = 0.3$\\
    $P(fr^c | f) = 0.7$\\
    $P(fr | f^c) = 0.25$\\
    $P(fr^c | f^c) = 0.75$\\

    Draw the probability tree to help solve the problem

    \part

    $$
    P(f \cap fr) = P(f) \cdot P(fr | f) = 0.6 \cdot \frac{1}{3} = 0.2
    $$

    \part

    $$P(fr) = 0.2 + (0.4 \cdot 0.25) = 0.2 + 0.1 = 0.3$$

    \part

    $$P(f | fr) = \frac{P(f \cap fr)}{P(fr)} = \frac{0.2}{0.3}$$

\end{homeworkProblem}
\pagebreak

\begin{homeworkProblem}
Of the patients in an emergency unit, 30\% have a sports injury. Pass records
for this emergency unit suggest that :\\

A patient with sports injury has a probability of 0.2 of being admitted to hospital\\
\\
A patient who doesn’t have a sports injury has a probability of 0.4 of being admitted to
hospital\\
\\
Now, a patient is chosen at random from those in the emergency unit.\\

    \begin{enumerate}[label=(\alph*)]
        \item What is the probability  that the chosen patient has a sports injury
        \item What is the probability that the chosen patient has a sports injury and is admitted to hospital
        \item Show that the probability that the chosen patient is admitted to hospital is 0.34
        \item Given that the chosen patient is admitted to hospital, find the conditional probability that the patient has a sports injury
    \end{enumerate}

    \solution

    Given:\\
    $P(\text{sports injury}) = 0.3$\\
    $P(\text{sports injury}^c) = 0.7$\\
    $P(\text{admitted} | \text{sports injury}) = 0.2$\\
    $P(\text{admitted} | \text{sports injury}^c) = 0.4$\\

    \part

    $$P(\text{sports injury}) = 0.3$$

    \part

    $$P(\text{sports injury} \cap \text{admitted}) = P(\text{sports injury}) \cdot P(\text{admitted} | \text{sports injury}) = 0.3 \cdot 0.2 = 0.06$$

    \part

    $$P(\text{admitted}) = P(\text{sports injury} \cap \text{admitted}) + P(\text{sports injury}^c \cap \text{admitted}) = 0.3 \cdot 0.2 + 0.7 \cdot 0.4 = 0.34$$

    \part

    $$P(\text{sports injury} | \text{admitted}) = \frac{P(\text{sports injury} \cap \text{admitted})}{P(\text{admitted})} = \frac{0.06}{0.34} = 0.17647 \approx 0.18$$

\end{homeworkProblem}
\pagebreak

\begin{homeworkProblem}

Darren and Charles are friends who often go to the cinema together. On such visits there
is a probability of 0.4 that Darren will buy popcorn.
The probability that Charles will buy
popcorn is 0.7 if Darren buys popcorn and 0.35 if he doesn’t.

When Darren and Charles visit the cinema together:
    \begin{enumerate}[label=(\alph*)]
        \item Find the probability that both buy popcorn
        \item Show that the probability that neither buys popcorn is 0.39
        \item Find the probability that exactly one of them buys popcorn
    \end{enumerate}
Sarah sometimes join Darren and Charles on their cinema visits. On these occasions,
the probability that sarah buys popcorn is 0.55 if both
of charles and darren buy popcorn and 0.25 if exactly one of darren and charles buys popcorn

When Darren, Sarah and Charles visit the cinema together:

    \begin{enumerate}[label=(\alph*)]
        \item What is the probability that the three of them buy popcorn
        \item Charles and sarah buy popcorn but Daren Doesn't
    \end{enumerate}

    \solution

    Given:\\
    $P(\text{Darren buys popcorn}) = 0.4$\\
    $P(\text{Charles buys popcorn} | \text{Darren buys popcorn}) = 0.7$\\
    $P(\text{Charles buys popcorn} | \text{Darren doesn't buy popcorn}) = 0.35$\\
    $P(\text{Darren doesn't buy popcorn}) = 0.6$\\

    \part

    $$
    \begin{align}
       & P(\text{Darren buys popcorn} \cap \text{Charles buys popcorn}) \\
        &= P(\text{Darren buys popcorn}) \cdot P(\text{Charles buys popcorn} | \text{Darren buys popcorn})\\
                                                                       &= 0.4 \cdot 0.7\\
                                                                       &= 0.28
    \end{align}
    $$

    \part

    $$
    \begin{align}
        & P(\text{Darren doesn't buy popcorn} \cap \text{Charles doesn't buy popcorn}) \\
        &= P(\text{Darren doesn't buy popcorn}) \cdot P(\text{Charles doesn't buy popcorn} | \text{Darren doesn't buy popcorn})\\
                                                                                                    &= 0.6 \cdot 0.65\\
                                                                                                    &= 0.39
    \end{align}
    $$


    \part

    $$
    \begin{align}
        & P(\text{Darren buys popcorn} \cap \text{Charles doesn't buy popcorn} \text{ or } \text{Charles buys popcorn} \cap \text{Darren doesn't buy popcorn}) \\
        &= P(\text{Darren buys popcorn}) \cdot P(\text{Charles doesn't buy popcorn} | \text{Darren buys popcorn}) \\
        &+ P(\text{Charles buys popcorn}) \cdot P(\text{Darren doesn't buy popcorn} | \text{Charles buys popcorn})\\
        &= 0.4 \cdot 0.3 + 0.6 \cdot 0.35\\
        &= 0.33
    \end{align}
    $$

    \part

    Given:
    $P(\text{Sarah buys popcorn} | \text{Darren buys popcorn} \cap \text{Charles buys popcorn}) = 0.55$\\
    $P(\text{Sarah buys popcorn} | \text{Darren buys popcorn} \cap \text{Charles doesn't buy popcorn}) = 0.25$\\
    $P(\text{Sarah buys popcorn} | \text{Darren doesn't buy popcorn} \cap \text{Charles buys popcorn}) = 0.25$\\

    $$
    \begin{align}
    P(\text{3 buy popcorn}) &= P(\text{Darren buys popcorn} \cap \text{Charles buys popcorn} \cap \text{Sarah buys popcorn}) \\
                            &= 0.4 \cdot 0.7 \cdot 0.55\\
                            &= 0.154
    \end{align}
    $$

    \part

    $$
    \begin{align}
        P(\text{Charles buys popcorn} \cap \text{Sarah buys popcorn} \cap \text{Darren doesn't buy popcorn}) &= 0.6 \cdot 0.35 \cdot 0.25\\
                                                                                                        &= 0.0525
    \end{align}
    $$


\end{homeworkProblem}
\pagebreak

\begin{homeworkProblem}


Following a flood, 120 tins were recovered from Lee’s corner shop.
Unfortunately, the water had washed off all the labels.
Of the tins, 50 contained pet food, 20 contained peas, 35
contained beans and the rest contained soup.

    \begin{enumerate}[label=(\alph*)]
\item Lee selects a tin at random, what is the probability that it contains soup?
\item Lee selects a tin at random, what is the probability that it doesn’t contain pet food
\item Lee selects two tins at random (without replacement), what is the probability that both
 contains peas
\item Lee selects two tins at random (without replacement), what is the probability that one
contains pet food and the other contains peas
\item Lee selects 3 tins at random (without replacement). What is the probability that one
contains pet food, one contains peas and one contains beans.
\item Find the probability that Lee will have to open more than two tins before he finds one
that doesn’t contain pet food.
    \end{enumerate}

    \solution

    \part

    Soup left: 120 - 50 - 20 - 35 = 15\\

    $$P(\text{soup}) = \frac{15}{120} = 0.125$$

    \part

    $$P(\text{pet food}^c) = \frac{120-50}{120} = 0.58333$$

    \part

    $$P(\text{peas} \cap \text{peas}) = \frac{20}{120} \cdot \frac{19}{119} = 0.00254$$

    \part

    $$P(\text{pet food} \cap \text{peas}) = \frac{50}{120} \cdot \frac{20}{119} \cdot 2 = 0.140056$$

    \part

    $$P(\text{pet food} \cap \text{peas} \cap \text{beans}) = \frac{50}{120} \cdot \frac{20}{119} \cdot \frac{35}{118} \cdot 3! = 0.12462612164 \approx 0.125$$

    \part

    What are the chances that it's just

    $$\frac{50}{120} \cdot \frac{49}{119} = 0.1715686 \approx 0.172$$



\end{homeworkProblem}
\pagebreak

\begin{homeworkProblem}
    A school employs 75 teachers. The following table summarizs their length of service at the school, classified by gender.

    \begin{tabular}{|c|c|c|c|}
        \hline
        & $<$ 3 years & 3 years to 8 years & $>$ 8 years\\
        \hline
        Female & 12 & 20 & 13\\
        \hline
        Male & 8 & 15 & 7\\
        \hline
    \end{tabular}
\\
    \begin{enumerate}[label=(\alph*)]
        \item Find the probability that a randomly selected teacher is a female
        \item Find the probability that a randomly selected teacher is female given that the teacher has more than 8 years service
        \item Find the probability that a randomly selected teacher is female given that the teacher has less than 3 years service
        \item State, giving a reason, whether or not the event of selecting a female teacher is independent of the event of selecting a teacher with less than 3 years service
    \end{enumerate}

    \solution

    \part

    $$P(\text{female}) = \frac{12+20+13}{75} = \frac{45}{75} = 0.6$$

    \part

    $$P(\text{female} | >\text{8 years}) = \frac{P(\text{female} \cap >\text{8 years})}{P(>\text{8 years})} = \frac{\frac{13}{75}}{\frac{20}{75}} = 0.65$$

    \part

    $$P(\text{female}|< \text{3 years}) = \frac{P(\text{female} \cap < \text{3 years})}{P(< \text{3 years})} = \frac{\frac{12}{75}}{\frac{20}{75}} = 0.6$$

    \part

    Test for independence: $$P(\text{female}| < \text{3 years}) = P(\text{female})$$\\
    Since both are 0.6, the events are independent.

\end{homeworkProblem}
\pagebreak

\begin{homeworkProblem}

A group of students bought, in total, 25 items of clothing at two shops: Mango and
Iora. The following table shows how many tops, jeans and sweaters were
bought at each of the two shops:\\
    \begin{tabular}{|c|c|c|c|}
        \hline
        & Top & Jeans & Sweaters\\
        \hline
        Mango & 3 & 7 & 5\\
        \hline
        Iora & 2 & 5 & 3\\
        \hline
    \end{tabular}
    \\
    One item of clothing is chosen at random from these 25 items:\\
    \begin{enumerate}[label=(\alph*)]
        \item What is the probability that the chosen item is a top?
        \item What is the probability that the chosen item was bought from Iora?
        \item What is the probability that the chosen item is a top and was bought from Iora?
        \item State with a reason whether the events (chosen item is a top) and (chosen item was bought from iora) are independent.
        \item Given that the chosen item is not a top, find the conditional probability that it was
    \end{enumerate}
    \solution

    \begin{tabular}{|c|c|c|c|c|}
        \hline
        & Top & Jeans & Sweaters & Total\\
        \hline
        Mango & 3 & 7 & 5 & 15\\
        \hline
        Iora & 2 & 5 & 3 & 10\\
        \hline
        Total & 5 & 12 & 8 & 25\\
        \hline
    \end{tabular}
    \\\\
    \part

    $$P(\text{top}) = \frac{5}{25} = 0.2$$

    \part

    $$P(\text{Iora}) = \frac{10}{25} = 0.4$$

    \part

    $$P(\text{top} \cap \text{Iora}) = \frac{2}{25} = 0.08$$

    \part

    $$P(\text{top} | \text{Iora}) = \frac{P(\text{top} \cap \text{Iora})}{P(\text{Iora})} = \frac{0.08}{0.4} = 0.2$$

    Test for independence: $$P(\text{top}) = P(\text{top} | \text{Iora})$$\\
    Since both are 0.2, the events are independent.
    \\\\
    \part

    $$P(\text{Iora} | \text{top}^c) = \frac{P(\text{Iora} \cap \text{top}^c)}{P(\text{top}^c)} = \frac{0.4 \cdot 0.8}{0.8} = 0.4$$



\end{homeworkProblem}
\pagebreak

\begin{homeworkProblem}
A volunteer for the Health Literacy Center was investigating the attitudes of students towards
smoking on campus. A random sample of 730 students from all four grade levels was taken.
Each student was given the statement "Smoking is dangerous for your health" and asked
whether they agreed, had no opinion or disagreed. The following contingency table summarizes the results.

    \begin{enumerate}[label=(\alph*)]
    \item Complete the following table knowing that
    \begin{enumerate}
        \item 387 Male students  participated
        \item 277 students  agreed and 97 students had no opinion
        \item 154 from those who agreed were female
        \item Only 36 from the males who participated had no opinion
    \end{enumerate}
    \item Among the participants, what is the probability that a student agreed P (A)?
    \item Show whether the events of students agreeing(A) and being a female student (F) are independent
    \item What is the probability that a student is a female given that the student had no opinion $P(F|N)$? What is the probability that a student disagreed, given that the student is Male $P(D|M)$?
    \item Use Bayes' Theorem to calculate the probability $P(M|A)$
    \end{enumerate}

    \solution\\

    \part

    \begin{tabular}{|c|c|c|c|c|}
        \hline
        & Agree (A) & No Opinion (N) & Disagree (D) & Totals\\
        \hline
        Male (M) & 123 & 36 & 228 & 387 \\
        \hline
        Female (F) & 154 &61 &128 &343 \\
        \hline
        Totals &277 & 97 & 356 &730 \\
        \hline
    \end{tabular}
    \\\\
    \part

    $$P(A) = \frac{277}{730} = 0.38$$

    \part

    $$P(A|F) = \frac{P(A \cap F)}{P(F)} = \frac{0.21}{0.47} = 0.45$$\\
    Test for independence: $$P(A) = P(A|F)$$\\
    Since $P(A) = 0.38$ and $P(A|F) = 0.45$, the events are not independent.

    \part

    $$P(F|N) = \frac{P(F \cap N)}{P(N)} = \frac{\frac{61}{730}}{\frac{97}{730}} = 0.63$$

    $$P(D|M) = \frac{P(D \cap M)}{P(M)} = \frac{\frac{228}{730}}{\frac{387}{730}} = 0.59$$

    \part

    $$P(M|A) = \frac{P(M \cap A)}{P(A)} = \frac{\frac{123}{730}}{\frac{277}{730}} = 0.44$$

\end{homeworkProblem}
\pagebreak

\begin{homeworkProblem}
Letter A is sent by first-class post and has a probability of 0.9 of being delivered next day. Letter B is sent by second-class post and has a probability of only 0.3 if being delivered the next day.
    \begin{enumerate}[label=(\alph*)]
        \item Find the probability that both letters are delivered the next day
        \item Find the probability that neither letter is delivered the next day
        \item Find the probability that at least one of the letters is delivered the next day
        \item Given that at least one of the letters is delivered the next day, find the probability that letter A is delivered the next day
    \end{enumerate}

    \solution\\

    Draw the Probability tree diagram to solve the problem.

    \part

    $$P(A \cap B) = P(A) \cdot P(B) = 0.9 \cdot 0.3 = 0.27$$

    \part

    $$P(A^c \cap B^c) = P(A^c) \cdot P(B^c) = 0.1 \cdot 0.7 = 0.07$$

    \part

    $$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.9 + 0.3 - 0.27 = 0.93$$
    OR, Using a tree diagram:\\
    $$P(A \cup B) = 0.9 \cdot 0.3 + 0.9 \cdot 0.7 + 0.1 \cdot 0.3 = 0.93$$

    \part

    $$P(a|\text{at least 1 letter}) = \frac{P(a \cap \text{at least 1 letter})}{P(\text{at least 1 letter})} = \frac{0.9 \cdot 0.3 + 0.9 \cdot 0.7}{0.93} = 0.9677419 \approx 0.9677$$

\end{homeworkProblem}
\pagebreak

\begin{homeworkProblem}
    \textbf{ANSWERS}
    \begin{enumerate}
        \item (b) $\frac{3}{4}$, $c = \frac{1}{2}$, classical interpretation
        \item $\frac{7}{15}$, $\frac{77}{150}$, $\frac{8}{15}$, Relative frequency interpretation
        \item 0.9
        \item 0.882, 0.1694
        \item 0.2,0.3, 0.2/0.3
        \item 0.3, 0.06, 0.34, 0.17647
        \item 0.28, 0.39, 0.33,0.154, 0.0525
        \item 0.125, 0.0583, 0.0266, 0.140, 0.125, 0.172
        \item 0.6, 0.65, 0.6, independent
        \item $\frac{1}{5}$, $\frac{2}{5}$, $\frac{2}{25}$, independent, $\frac{2}{5}$
        \item B)P(A) = $\frac{377}{730}$, C) not Independent, D) $\frac{61}{97}, \frac{228}{387}$, E) 0.44404
        \item A) 0.27, B)0.07, C)0.93, D)0.9677
    \end{enumerate}

    11a)

    \begin{tabular}{|c|c|c|c|c|}
        \hline
        & Agree (A) & No Opinion (N) & Disagree (D) & Totals\\
        \hline
        Male (M) & 123 & 36 & 228 & 387 \\
        \hline
        Female (F) & 154 &61 &128 &343 \\
        \hline
        Totals &277 & 97 & 356 &730 \\
        \hline
    \end{tabular}
\end{homeworkProblem}

\end{document}
