Problem 1.

a)

The sample space \(S = \{H, D, C, S\}\) is discrete (and finite). An event could be \[A = \{\text{the card is red}\} = \{H, D\}\] #### b)

The sample space \(S = \{0, 1, 2, 3, \cdots 16\}\) is discrete (and finite). An event could be

\[A = \{\text{the Cardinals win an odd number of games}\} = \{1, 3, 5, 7, 9, 11, 13, 15\}\] or \[B = \{\text{the Cardinals win at most 3 games}\} = \{0,1,2,3\}\]

c)

The sample space \(S = \{0,1,2,3,4,\cdots\}\) is discrete (but not finite). An event could be \[A = \{\text{the account has at least 100 followers}\} = \{100, 101, 102, 103, \cdots\} = \{n| n\in \mathbb N, \ and \ n \geq 100\}\]

d)

The sample space \(S = (0, \infty)\) is continuous. An event could be \[A = \{\text{Phelps breaks his own personal record}\} = (0, 111.51)\] where the unit of measurement is seconds. Another event could be \[A^c = \{\text{Phelps fails to break his personal record}\} = [111.51, \infty).\]

Problem 2.

a)

\[A - B = A \cap B^c\]

\[A \triangle B = (A\cup B) - (A\cap B) = (A \cup B) \cap (A\cap B)^c\]

Problem 3.

Ask me if you have questions for this one.

Problem 4.

  1. Using a standard 52 card deck, how many ways can 13 cards be dealt to each of four different players?

\[\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}\]

  1. Using a standard 52 card deck, how many ways can a 5 card poker hand be dealt? How many of these hands will be a flush? a flush consists of five cards of the same suit.

Total number of hands:

\[\binom{52}{5} = 2598960\] Flushes: Choose suit first (\(4\) ways). Then choose \(5\) chards.

\[4\binom{13}{5} = 5148\]

  1. How many ways can two octopi shake hands if a hand shake consists of each octopus using a single tentacle?

\[\binom{8}{1}\binom{8}{1}\]

  1. How many ways can two octopi shake hands if a hand shake consists of each octopus using two tentacles? for example, two humans can shake hands (using both hands) in just two ways: (right ↔ right, lef t ↔ lef t) or (lef t ↔ right, right ↔ lef t).

Two ways (at least) to think about this problem. First method: (i) Choose 2 tentacles from first octopus (call them A and B), (ii) 8 choices to shake with tentacle A (iii) 7 choices to shake with tentacle B.

\[\binom{8}{2}\cdot 8 \cdot 7 = 1568\]

Second method: (i) Choose 2 tentacles for first octupus (call them A and B) (ii) Choose 2 tentacles from second octopus (call them B and C) (iii) 2 ways to do a hand shake: (A to C and B to D) or (A to D and B to C)

\[\binom{8}{2}\binom{8}{2}\cdot 2 = 1568\]

Problem 5.

Recall the Hypergeometric formula:

\[P(\text{exactly x rabbits in the sample are tagged}) = \frac{\binom{k}{x}\binom{N-k}{n-x}}{\binom{N}{n}}\]

In this problem, we have \(k=10\), \(n=20\), \(x=4\) and \(N\) is unknown. So

\[P(\text{exactly $4$ rabbits in the sample are tagged}) = \frac{\binom{10}{4}\binom{N-10}{16}}{\binom{N}{20}},\]

which can be computed easily in R

N <- 30:120
prob <- choose(10, 4)*choose(N-10, 16)/choose(N, 20)
plot(N, prob, pch=21, cex=1.2, bg='orange')

#Find which N leads to highest probability
best <- which.max(prob)
N[best]
## [1] 49
#Find corresponding highest probability
prob[best]
## [1] 0.2800586

So it is “most likely” that \(N=49\).

Problem 6.

Note: I am giving more detail here than I expect from you. But if you are confused, try to read this in detail.

The outcomes in the sample space represent a selection of \(7\) books. Hence there are \[|S| = \binom{23}{7} = 245157\] total possible outcomes.

The event in question is \[A = \{\text{at least $2$ books in each subject}\}.\]

We can write \(A\) as the union of three disjoint sets: \[\begin{align*} E_1 = \{\text{$3$ chem books, $2$ math books and $2$ astr. books}\} \\[1.2ex] E_2 = \{\text{$2$ chem books, $3$ math books and $2$ astr. books}\} \\[1.2ex] E_3 = \{\text{$2$ chem books, $2$ math books and $3$ astr. books}\} \end{align*}\]

To clarify, the following things are true:

  • If \(A\) occurs, then at least one of the events \(E_1\), \(E_2\) or \(E_3\) occurs. This means that \(A \subset (E_1\cup E_2 \cup E_3)\).
  • If any of the events \(E_1\), \(E_2\) or \(E_3\) occurs, then \(A\) must also occur. This means that \((E_1\cup E_2 \cup E_3) \subset A\).
  • The two points above impliy that \(A= (E_1\cup E_2 \cup E_3)\).
  • If \(E_1\) occurs, then \(E_2\) and \(E_3\) must not occur. Same goes for \(E_2\) and \(E_3\). Thus the events \(E_1\), \(E_2\) and \(E_3\) are pairwise disjoint.

Since \(A = (E_1\cup E_2 \cup E_3)\), we can write \[|A| = |(E_1\cup E_2 \cup E_3)|,\] and since the events \(E_1, E_2\) and \(E_3\) are pairwise disjoint, we can write:

\[|A| = |E_1| + |E_2| + |E_3| = \binom{10}{3}\binom{8}{2}\binom{5}{2} +\binom{10}{2}\binom{8}{3}\binom{5}{2} +\binom{10}{2}\binom{8}{2}\binom{5}{3} = 71400.\] So the probability of the event in question can be found by \[P(A) = \frac{|A|}{|S|} = \frac{71400}{245157} = 0.2912.\]

Problem 7.

There are \(2^4=16\) possible outcomes, and the sample space can be written as

\[S = \{0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111,1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111\}.\]

Now look for outcomes whichs lead to event \(A\) “occuring” \[A = \{0011, 0110, 0111, 1011, 1100, 1101, 1110, 1111\}.\]

So we have

\[P(A) = \frac{|A|}{|S|} = \frac{8}{16} = 0.5\]