Question 1.3.1
Jane must take two tests, call them \(T_1\) and \(T_2\). The probability that she passes test \(T_1\) is 0.8, that she passes test \(T_2\) is 0.7, and that of passing both tests is 0.6.
Calculate the probability that:
She passes at least one test.
She passes at most one test.
She fails both tests.
She passes only one test.
Let: \[\begin{aligned}
A &= \{ \text{Jane passes test $T_1$} \} \\
B &= \{ \text{Jane passes test $T_2$} \}\end{aligned}\]
We are told that \[\mathbb{P}\left( A\right) =0.80,
\mathbb{P}\left( B\right) =0.70, \text{ \ and
\ }\mathbb{P}\left( A\cap B\right) =0.6.\]
We note that \[\left\{ \mbox{Passes at least one test}\right\} = A \cup B.\]
Hence, \[\begin{aligned}
\mathbb{P}\left( A\cup B\right)
& =
\mathbb{P}\left( A\right) +\mathbb{P}\left( B\right) -\mathbb{P}\left( A\cap B\right) \\
&=
0.80+0.70-0.60 \\
& = 0.90.
\end{aligned}\]
The probability that Jane passes at least 1 test is 0.9.
We notice that \[\begin{aligned}
\{ \text{Passes at most one test}\}
&=
\left\{ \text{Fails at least one test}\right\} \\
&=
A^{c}\cup B^{c}.\end{aligned}\]
And we notice \[A^{c}\cup B^{c} = (A\cap B) ^{c}\text{ \ \ De Morgan rule\ \ }\] which gives one way of computing its probability.
\[\begin{aligned}
\mathbb{P}( \{ \text{Passes at most one test} \} )
&= \mathbb{P}\left(\left( A\cap B\right) ^{c}\right) \text{ \ \ \ } \\
&=1-\mathbb{P}\left( A\cap B\right) \\
&=1-0.60=0.40.\end{aligned}\]
The probability that Jane passes at most 1 test is 0.4.
Similarly, we observe by De Morgan rule, \[\begin{aligned}
\{ \text{Fails both tests}\}
&=
A^{c}\cap B^{c} \\
&=
( A\cup B) ^{c} \end{aligned}\] which gives one way of computing its probability.
\[\begin{aligned}
\mathbb{P}( \{ \text{Fails both tests} \} )
&=
\mathbb{P}\left( \left( A\cup B\right) ^{c}\right) \\
&=1-\mathbb{P}\left( A\cup B\right) \\
&=1-0.90 \\
&=
0.10 \text{ from Part (a)}.\end{aligned}\]
Therefore, the probability that Jane fails both tests is 0.1.
We first decompose the event into the union of two disjoint events: \[\left\{ \text{Passes only one test}\right\} =%
\left( A\cap B^{c}\right)
\cup
\left( A^{c}\cap B\right)\]
We hence have \[
\begin{aligned}
\mathbb{P}(\left\{ \text{Passes only one test}\right\})
&=
\mathbb{P}\left( A\cap B^{c}\right)
+
\mathbb{P}\left( A^{c}\cap B\right) \\
&=
\left[ \mathbb{P}\left( A\right) -\mathbb{P}\left( A\cap B\right) \right]
+ \left[ \mathbb{P}\left( B\right) -\mathbb{P}\left( A\cap B\right) \right] \\
&=
\mathbb{P}\left( A\right) +\mathbb{P}\left( B\right) -2 \times \mathbb{P}\left( A\cap B\right) \\
&=
0.80+0.70-2\times 0.60 =
0.30.
\end{aligned}
\]
The probability that Jane passes only 1 test is 0.3.
Question 1.3.2
Prove the Bonferroni inequality: \[\mathbb{P}\left( \bigcap _{i=1}^{n}A_{i}\right) \geq \sum_{i=1}^{n} \mathbb{P}\left( A_{i}\right) -\left( n-1\right)\]
Show every step of the proof. List any results, rules, or theorems you use in the respective lines of your proof.
Hint: \[\bigcap _{i=1}^{n}A_{i} \, = \,
\left( \bigcup _{i=1}^{n}A_{i}^{c}\right) ^{c}.\]
\[
\begin{aligned}
\mathbb{P}\left( \bigcap _{i=1}^{n}A_{i}\right) & =\mathbb{P}\left( \left( \bigcup
_{i=1}^{n}A_{i}^{c}\right) ^{c}\right) \\
& =1-\mathbb{P}\left( \bigcup _{i=1}^{n}A_{i}^{c}\right) &\text{ \ complement
rule}\\
& \geq 1-\sum_{i=1}^{n}\mathbb{P}\left( A_{i}^{c}\right) &\text{ \ Boole's inequality}\\
& =1-\sum_{i=1}^{n}\left( 1-\mathbb{P}\left( A_{i}\right) \right) &\text{ \ complement
rule} \\
& =1-\sum_{i=1}^{n}1+\sum_{i=1}^{n}\mathbb{P}\left( A_{i}\right) \\
&=1-n+\sum_{i=1}^{n}\mathbb{P}\left( A_{i}\right) \\
& =\sum_{i=1}^{n}\mathbb{P}\left( A_{i}\right) -\left( n-1\right) .
\end{aligned}
\]
Question 1.3.3
Suppose that Al watches the six o’clock news \(\frac{2}{3}\) of the time, watches the eleven o’clock news \(\frac{1}{2}\) of the time, and watches both the six o’clock and eleven o’clock news \(\frac{1}{3}\) of the time. For a randomly selected day, what is the probability that Al watches only the six o’clock news? For a randomly selected day, what is the probability that Al watches neither news?
Let \(A\) be the event “Al watches the six o’clock news” and \(B\) be the event “Al watches the eleven o’clock news.” Then \(\mathbb{P}(A) = \frac{2}{3}\), \(\mathbb{P}(B) = \frac{1}{2}\), and \(\mathbb{P}(A \cap B) = \frac{1}{3}\).
Only the six o’clock news: This is the event \(A \setminus (A \cap B)\), so \[
\begin{aligned}
\mathbb{P}(A \setminus (A \cap B)) &= \mathbb{P}(A) - \mathbb{P}(A \cap B) \\
&= \frac{2}{3} - \frac{1}{3} \\
&= \frac{1}{3}
\end{aligned}
\]
Neither news: This is the event \((A \cup B)^c\), so \[
\begin{aligned}
\mathbb{P}((A \cup B)^c) &= 1 - \mathbb{P}(A \cup B) \\
&= 1 - \left[\mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A \cap B)\right] \\
&= 1 - \left[\frac{2}{3} + \frac{1}{2} - \frac{1}{3}\right] \\
&= 1 - \frac{5}{6} \\
&= \frac{1}{6}
\end{aligned}
\]
Therefore, the probability that Al watches only the six o’clock news is \(\boldsymbol{\dfrac{1}{3}}\), and the probability that Al watches neither news is \(\boldsymbol{\dfrac{1}{6}}\).
Question 1.3.4
Suppose that an employee arrives late 10% of the time, leaves early 20% of the time, and both arrives late and leaves early 5% of the time. What is the probability that on a given day that employee will either arrive late or leave early (or both)?
Let \(L\) be the event “arrives late” and \(E\) be the event “leaves early,” so \(\mathbb{P}(L) = 0.10\), \(\mathbb{P}(E) = 0.20\), and \(\mathbb{P}(L \cap E) = 0.05\). By the principle of inclusion–exclusion (Theorem 1.3.3): \[
\begin{aligned}
\mathbb{P}(L \cup E) &= \mathbb{P}(L) + \mathbb{P}(E) - \mathbb{P}(L \cap E) \\
&= 0.10 + 0.20 - 0.05 \\
&= 0.25
\end{aligned}
\] Therefore, the probability that the employee either arrives late or leaves early (or both) is \(\boldsymbol{25\%}\).
Question 1.3.5
Suppose a fair coin is tossed five times in a row.
What is the probability of getting all five heads?
What is the probability of getting at least one tail?
There are \(2^5 = 32\) equally likely possible outcomes, and only one of them is all five heads. Thus, \[
\mathbb{P}(\text{all five heads}) = \frac{1}{32} = 0.03125
\] Therefore, the probability of getting all five heads is \(\boldsymbol{\dfrac{1}{32}}\).
Let \(B\) be the event of getting at least one tail, and let \(A\) be the event of getting all five heads. There will be at least one tail unless all five coins are heads, so \(B = A^c\). Thus, \[
\begin{aligned}
\mathbb{P}(B) &= \mathbb{P}(A^c) \\
&= 1 - \mathbb{P}(A) \\
&= 1 - \frac{1}{32} \\
&= \frac{31}{32} \\
&= 0.96875
\end{aligned}
\] Therefore, the probability of getting at least one tail is \(\boldsymbol{\dfrac{31}{32}}\).
Question 1.3.6
Suppose a card is chosen uniformly at random from a standard 52-card deck.
What is the probability that the card is a jack?
What is the probability that the card is a club?
What is the probability that the card is both a jack and a club?
What is the probability that the card is either a jack or a club (or both)?
There are only 4 jacks in a standard 52-card deck, so \[
\mathbb{P}(\text{jack}) = \frac{4}{52} = \frac{1}{13} \approx 0.0769
\] Therefore, the probability that the card is a jack is \(\boldsymbol{\dfrac{1}{13}}\).
There are 13 clubs in the deck, so \[
\mathbb{P}(\text{club}) = \frac{13}{52} = \frac{1}{4} = 0.25
\] Therefore, the probability that the card is a club is \(\boldsymbol{\dfrac{1}{4}}\).
There is only one card that is both a jack and a club (the jack of clubs), so \[
\mathbb{P}(\text{jack and club}) = \frac{1}{52} \approx 0.01923
\] Therefore, the probability that the card is both a jack and a club is \(\boldsymbol{\dfrac{1}{52}}\).
By the principle of inclusion–exclusion (Theorem 1.3.3), using the results from parts (a)–(c): \[
\begin{aligned}
\mathbb{P}(\text{jack or club}) &= \mathbb{P}(\text{jack}) + \mathbb{P}(\text{club}) - \mathbb{P}(\text{jack and club}) \\
&= \frac{4}{52} + \frac{13}{52} - \frac{1}{52} \\
&= \frac{16}{52} \\
&= \frac{4}{13} \\
&\approx 0.3077
\end{aligned}
\] Therefore, the probability that the card is either a jack or a club (or both) is \(\boldsymbol{\dfrac{4}{13}}\).