(a) Toss a coin four times.

The sample space can be represented by listing the possible outcomes: \(\{HHHH\}, \{HHHT\}, \{HHTH\}, \{HTHH\}, \\ \{THHH\}, \{HHTT\}, \{HTHT\}, \{HTTH\}, \\ \{THTH\}, \{TTHH\}, \{THHT\}, \{TTTH\}, \\ \{TTHT\}, \{THTT\}, \{HTTT\}, \{TTTT\}\)

(b) Count the number of insect-damaged leaves on a plant

Since we are only interested in the number, and not in the extent of damage, the sample space is the set of nonnegative integers: \(S=\{0,1,2,3,...\}\). If we knew the total number of leaves (unlikely), we could put an upper bound on the possible outcomes.

(c) Measure the lifetime (in hours) of a particular brand of light bulb

The lifetime of a lightbulb can’t be negative, but there’s no reason the answer needs to be a whole number. There’s also no clear upper bound. So we can say \(S=[0,\infty]\).

(d) Record the weights of 10-day-old rats

This is similar to the lighbulb problem. The weights don’t need to be in whole numbers. While there is no need to impose an upper bound, it is easier to find one here than it was in the lighbulb problem. We would be very surprised to find a 10-day-old rat weighing more than, say, 31 oz. So, measuring in ounces, we can say \(S=[0,32]\) (though any upper bound will do)

(e) Observe the proprtion of defectives in a shipment of electronic components.

The sample space here will take on rational number values. Suppose there are \(N\) total components. Then the possible values for the proportion of defectives are \(S=\left\{\frac{1}{N}, \frac{2}{N},...,\frac{N}{N}\right\}\).

(a)

\(A \setminus B = A \setminus (A\cup B)=A\cap B^c\)

Proof (i = iii):

Note that, to prove set equality, we prove (i) is a subset of (iii) and (iii) is a subset of (i).

\[\begin{align*} x \in A \setminus B &\Rightarrow x\in A \text{ and } x \notin B \\ &\Rightarrow x \in A \text{ and } x \in B^c \\ &\Rightarrow x \in A \cap B^c \\ &\Rightarrow A \setminus B \subseteq A \cap B^c \\ \\ x \in A\cap B^c &\Rightarrow x \in A \text{ and } x \in B^c \\ &\Rightarrow x \in A \text{ and } x \notin B \\ &\Rightarrow x \in A \setminus B \\ &\Rightarrow A \cap B^c \subseteq A \setminus B \\ &&\blacksquare \end{align*}\]

The proofs of the other identities follow from this because:

\[\begin{align*} A \setminus (A \cap B) &= A\cap (A \cap B)^c \\ &=A \cap (A^c \cup B^c) && \text{(DeMorgan's Laws)} \\ &=(A \cap A^c) \cup (A \cap B^c) && \text{(Distributive property)} \\ &= \emptyset \cup (A \cap B^c) \\ &= A\cap B^c = A \setminus B \\ && \blacksquare \end{align*} \]

We can verify some of this in R. We use the setdiff() and intersect() commands. We define a sample space S to contain A and B, so we can define \(B^c\) as \(S\setminus B\).

# The Sample Space
S=1:20
# Subsets A and B, corresponding to problem
A=1:10
B=6:10
B_complement = setdiff(S,B)
A_complement = setdiff(S,A)

# Set Difference
setdiff(A,B)

## [1] 1 2 3 4 5

# Set difference: A and A intersection B
setdiff(A, intersect(A,B))

## [1] 1 2 3 4 5

# A intersection B complement
intersect(A,B_complement)

## [1] 1 2 3 4 5

As you can see, we obtain the same answer from all three versions and, at least in this case, we have verified the identity.

(b)

\(B=(B \cap A) \cup (B \cap A^c)\)

Proof

Let \(S\) represent the sample space

\[ \begin{align*} B&=(B \cap A) \cup (B \cap A^c) \\ &= B \cap (A \cup A^c) && \text{(Set distribution)} \\ &= B \cap S && (A \cup A^c =S) \\ &= B && (B \cap S = B) \\ \\ && \blacksquare \end{align*} \]

We again verify in R, with A, B, and S as defined above. We add the union() command to our repertoire.

B

## [1]  6  7  8  9 10

# (B intersect A) U (B intersect A complement)
union(intersect(A,B),intersect(A_complement,B))

## [1]  6  7  8  9 10

Again, as expected, the identity holds in practice. We will dispense with the R examples in (c) and (d) as they are very similar.

(c)

$ B A = B A^c$

This was proven in part (a).

(d)

\(A \cup B = A \cup (B \cap A^c)\). The proof is evident if we note, as in (b), that \(A \cup A^c = S\). Then \(A \cup (B \cap A^c) = (A \cup B) \cap (A \cup A^c) = (A \cup B) \cap S = A \cup B\).

(a) Commutativity holds for intersection and union of sets A and B

\[ \begin{align*} x \in A \cup B &\Rightarrow x \in A \text{ or } x \in B \\ &\Rightarrow x \in B \text{ or } x \in A \\ &\Rightarrow x \in B \cup A \\ && \blacksquare \end{align*} \]

The argument for intersections is the same. We omit part (b) because the argument is essentially the same, with an extra step.

(c) Prove DeMorgan’s Laws for sets A and B.

In this case, we will only prove DeMorgan’s Law for set intersections. The proof for unions is, again, very similar.

\[ \begin{align*} x \in (A \cap B)^c &\Rightarrow x \notin (A \cap B) \\ &\Rightarrow x \notin A \text{ or } x \notin B \\ &\Rightarrow x \in A^c \text{ or } x \in B^c \\ &\Rightarrow x \in A^c \cup B^c \\ &\Rightarrow (A \cap B)^c \subseteq A^c \cup B^c \\ \\ x \in A^c \cup B^c &\Rightarrow x \notin A \text{ or } x \notin B \\ &\Rightarrow x \notin (A \cap B) \\ &\Rightarrow x \in (A \cap B)^c \\ &\Rightarrow A^c \cup B^c \subseteq (A \cap B)^c \\ && \blacksquare \end{align*} \]

(a) “Either A or B or both”

This is just inclusive or, \(A \cup B\). However, we don’t have the union operation available. However, theorem 1.2.9 (p. 10) states that \(P(A\cup B) = P(A)+P(B)-P(A \cap B)\), which does exclusively use the tools available to us.

(b) “Either A or B but not both”

This is exclusive or: we want the probability of \(A \cap B^c\) or \(B \cap A^c\). In other words, we’re interested in P((A B^c)(B A^c)). Because \((A \cap B^c) \cup (B \cap A^c) = \emptyset\), we have \(P((A \cap B^c) \cup (B \cap A^c)) = P(A \cap B^c)+P(B \cap A^c) \\ = [P(A)-P(A \cap B)]+[P(B)-P(A \cap B)] \\ = P(A)+P(B)-2P(A \cap B)\)

(c) “At least one of A or B”

This is the same as part (a).

(d) “At most one of A or B”

It’s easiest to think about this problem in terms of what it excludes: \(A\cap B\). Everything else is included: Just \(A\), just \(B\), and neither \(A\) nor \(B\). Recalling that the total probability is 1, we can represent the solution as \(1-P(A\cap B)\). This is the probability of the whole sample space minus the intersection of \(A \cap B\).

Part (a)

State, in words, the event \(A \cap B \cap C\)

Event C represents twin females; B represents identical twins; and C represents twins in general. The intersection of these three, then, represents identical female twins.

Part (b)

Find \(P(A\cap B\cap C)\)

We take \(P(C) \times P(B) \times P(A) = \frac{1}{90} \times \frac{1}{3} \times \frac{1}{2} = \frac{1}{540}\).