Back to the class Section 3.2 #20: A coin is tossed and a die is rolled. Find the probability of tossing a tail and then rolling a number greater than 2. Solution: Making a tree diagram for this problem yields the following picture:
From this we observe that the event being described is of size $4$ (it contains only the 3, 4, 5, and 6 below "T"). Moreover the size of the whole sample space is 12. Therefore the probability of this event is given by
$$P(\mathrm{event})=\dfrac{4}{12}=\dfrac{1}{3}=0.3333.$$
Section 3.2 #23: In a sample of 1000 U.S. adults, 180 dine out at a restaurant more than once per week. Two U.S. adults are selected at random without replacement.
a.) Find the probability that both adults dine out more than once per week.
b.) Find the probability that neither adult dines out more than once per week.
c.) Find the probabiltiy that at least one of the two adults dines out more than once per week.
d.) Which of the events can be considered unusual? Solution: We are told that the probability a single adult dines out per week is $\dfrac{180}{1000}=0.18$. For part a.), we are asked to find the probability of the event that both the first and second adult sampled dines out more than once per week: use the "addition rule" to compute
$$\begin{array}{ll}
P(\mathrm{event})&=P((\mathrm{first \hspace{2pt} dines \hspace{2pt} out}) \mathrm{AND}(\mathrm{second \hspace{2pt} dines \hspace{2pt} out})) \\
&=P(\mathrm{first \hspace{2pt} dines \hspace{2pt} out}) P(\mathrm{second \hspace{2pt} dines \hspace{2pt} out}|\mathrm{first \hspace{2pt} dines \hspace{2pt} out}) \\
&= \dfrac{180}{1000} \dfrac{179}{999} \\
&=0.0322.
\end{array}$$
For part b.), we are asked to find the probability that both of the adults do not dine out more than once per week, so use the addition rule to compute
$$\begin{array}{ll}
P((\mathrm{1st \hspace{2pt} doesn't \hspace{2pt} dine \hspace{2pt} out}) \mathrm{AND} (\mathrm{2nd \hspace{2pt} doesn't \hspace{2pt} dine \hspace{2pt} out})) &= P(\mathrm{1st \hspace{2pt} doesn't \hspace{2pt} dine \hspace{2pt} out}) P(\mathrm{2nd \hspace{2pt} doesn't \hspace{2pt} dine \hspace{2pt} out}|\mathrm{1st \hspace{2pt} doesn't \hspace{2pt} dine \hspace{2pt} out}) \\
&= \dfrac{820}{1000} \dfrac{819}{999} \\
&=0.6722.
\end{array}$$
For part c.), note that this event is the complement of the event in part b.), because "at least one" dining out more than once per week is "the opposite" of "none" dining out more than once per week. Hence we compute
$$P(\mathrm{none \hspace{2pt} dine \hspace{2pt} out})=1-P(\mathrm{1st \hspace{2pt} and \hspace{2pt} 2nd \hspace{2pt} dine \hspace{2pt} out}) =1-0.6722=0.3278.$$
Section 3.3 #18: You roll a die. Find each probability.
a.) Rolling a 5 or a number greater than 3.
b.) Rolling a number less than 4 or an even number.
c.) Rolling a 2 or an odd number. Solution: Note that there is a total sample size of 6 here. For part a.), use the addition and multiplication rules to compute
$$\begin{array}{ll}
P(\mathrm{rolling \hspace{2pt} a\hspace{2pt} 5\hspace{2pt}or\hspace{2pt}number\hspace{2pt}greater\hspace{2pt}than\hspace{2pt}3}) \\
=P(\mathrm{rolling\hspace{2pt}a\hspace{2pt}5})+P(\mathrm{rolling\hspace{2pt}a\hspace{2pt}number\hspace{2pt}greater\hspace{2pt}than\hspace{2pt}3})-P(\mathrm{rolling\hspace{2pt}a\hspace{2pt}5\hspace{2pt}and\hspace{2pt}rolling\hspace{2pt}a\hspace{2pt}number\hspace{2pt}greater\hspace{2pt}than\hspace{2pt}3}) \\
=\dfrac{1}{6}+\dfrac{3}{6}-P(\mathrm{rolling\hspace{2pt}a\hspace{2pt}5})P(\mathrm{rolling\hspace{2pt}a\hspace{2pt}number\hspace{2pt}greater\hspace{2pt}than\hspace{2pt}3|rolling\hspace{2pt}a\hspace{2pt}5}) \\
=\dfrac{1}{6} + \dfrac{3}{6} - \dfrac{1}{6}(1) \\
= \dfrac{3}{6} \\
=\dfrac{1}{2} \\
=0.5.
\end{array}$$
For part b.), use the addition and multiplication rules to compute
$$\begin{array}{ll}
P(\mathrm{rolling\hspace{2pt} a\hspace{2pt} number\hspace{2pt} less\hspace{2pt} than\hspace{2pt} 4\hspace{2pt} or\hspace{2pt} an\hspace{2pt} even\hspace{2pt} number}) \\
=P(\mathrm{rolling\hspace{2pt} a\hspace{2pt} number\hspace{2pt} less\hspace{2pt} than\hspace{2pt} 4}) + P(\mathrm{rolling\hspace{2pt} an\hspace{2pt} even}) - P(\mathrm{rolling\hspace{2pt} a\hspace{2pt} number\hspace{2pt} less\hspace{2pt} than\hspace{2pt} 4\hspace{2pt} and\hspace{2pt} rolling\hspace{2pt} an\hspace{2pt} even}) \\
=\dfrac{3}{6} + \dfrac{3}{6} - P(\mathrm{rolling\hspace{2pt} a\hspace{2pt} number\hspace{2pt} less\hspace{2pt} than\hspace{2pt} 4})P(\mathrm{rolling\hspace{2pt} an\hspace{2pt} even | rolling\hspace{2pt} a\hspace{2pt} number\hspace{2pt} less\hspace{2pt} than\hspace{2pt} 4}) \\
=1-\dfrac{3}{6}\dfrac{1}{3} \\
=\dfrac{5}{6} \\
=0.8333
\end{array}$$
For part c.), use the addition and multiplication rules to compute
$$\begin{array}{ll}
P(\mathrm{rolling\hspace{2pt}a\hspace{2pt}2\hspace{2pt}or\hspace{2pt}an\hspace{2pt}odd\hspace{2pt}number}) \\
=P(\mathrm{rolling\hspace{2pt}a\hspace{2pt}2})+P(\mathrm{rolling\hspace{2pt}an\hspace{2pt}odd})-P(\mathrm{rolling\hspace{2pt}a\hspace{2pt}2\hspace{2pt}and\hspace{2pt}rolling\hspace{2pt}an\hspace{2pt}odd}) \\
=\dfrac{1}{6}+\dfrac{3}{6}-P(\mathrm{rolling\hspace{2pt}a\hspace{2pt}2})P(\mathrm{rolling\hspace{2pt}an\hspace{2pt}odd|rolling\hspace{2pt}a\hspace{2pt}2}) \\
=\dfrac{1}{6}+\dfrac{3}{6}-\dfrac{1}{6}(0) \\
= \dfrac{4}{6} \\
=\dfrac{2}{3} \\
=0.6666
\end{array}$$
Section 4.1 #19: From the given data:
Television
0
1
2
3
Households
26
442
728
1404
a.) construct a probability distribution, and
b.) graph the distribution using a histogram. Solution: For part a.), first sum the number of households:
$$26+442+728+1404=2600.$$
Find the probabilities from the data:
Televisions
0
1
2
3
Probability
$\dfrac{26}{2600}=0.01$
$\dfrac{442}{2600}=0.17$
$\dfrac{728}{2600}=0.28$
$\dfrac{1404}{2600}=0.54$
For part b.):
Section 4.1 #21: Use the probability distribution in #19 to find the probability of randomly selecting a household that has...
a.) one or two televisions,
b.) two or more televisions, and
c.) between 1 and 3 televisions (inclusive). Solution: For part a.),
$$\begin{array}{ll}
P(\mathrm{one\hspace{2pt}or\hspace{2pt}two})&=P(\mathrm{one})+P(\mathrm{two})-P(\mathrm{one \hspace{2pt} and\hspace{2pt}two}) \\
&= 0.17 + 0.28 - P(\mathrm{one})P(\mathrm{two}|\mathrm{one}) \\
&= 0.17 + 0.28 - (0.17)(0) \\
&= 0.45.
\end{array}$$
For part b.), compute
$$\begin{array}{ll}
P(\mathrm{two\hspace{2pt}or\hspace{2pt}more})&=P(\mathrm{two})+P(\mathrm{more\hspace{2pt}than\hspace{2pt}2})-P(\mathrm{two\hspace{2pt}and\hspace{2pt}more\hspace{2pt}than\hspace{2pt}two}) \\
&= 0.28 + 0.54 - 0 \\
&=0.82.
\end{array}$$
For part c.), compute
$$\begin{array}{ll}
P(\mathrm{between\hspace{2pt}one\hspace{2pt}and\hspace{2pt}three\hspace{2pt}inclusive}) &=P(1)+P(2)+P(3)\\
&= 0.17+0.28+0.54 \\
&=0.99.
\end{array}$$
