Riemann surface
matlab

For each of the following experiments, describe a reasonable sample space:
(a) Toss a coin four times.
(b) Count the number of insect-damaged leaves on a plant.
(c) Measure the lifetime (in hours) of a particular brand of light bulb.
(d) Three people arrive at an airport checkpoint. Two of the three are randomly chosen to complete a survey.

Solutions

(a) Toss coin 4 times:
$${(H, H, H, H), \ldots}=\left{\left(x_1, x_2, x_3, x_4\right): x_i \in{H, T}\right}$$
or
$${0,1,2,3,4}$$
(b) Count number of insect-damaged leaves:
$$\begin{array}{cl} {0,1, \ldots, N} & N=# \text { leaves (or upper bound) } \ {0,1,2, \ldots} & \text { if no upper bound is available } \end{array}$$
1
Statistics STAT:5100 (22S:193), Fall 2015
Tierney
\begin{aligned} {0,1,2, \ldots} & \text { if rounded (can put in upper limit) } \ {[0, \infty) } & \text { if fractional hours are allowed } \end{aligned}
(d) Two out of three people chosen to complete a survey: Suppose the people are labeled $A, B$, and $C$. One possible sample space is is the collection of all subsets of size 2 that can be chosen from the set ${A, B, C}$ :
$${{A, B},{A, C},{B, C}}$$
Another possibility is the collection of all ordered pairs that can be formed:
$${(A, B),(B, A),(A, C),(C, A),(B, C),(C, B)}$$

The set-theoretic difference $A \backslash B=A \cap B^c$ is the set of all elements in $A$ that are not in $B$. The symmetric difference $A \Delta B=(A \backslash B) \cup(B \backslash A)$ is the set of all elements in either $A$ or $B$ but not both. Verify the following identities:
(a) $A \backslash B=A \backslash(A \cap B)$
(b) $A \Delta B=A^c \Delta B^c$
(c) $A \cup B=A \cup(B \backslash A)$
(d) $B=(B \cap A) \cup\left(B \cap A^c\right)$

(a) $A \backslash B$ is defined as $A \cap B^c$. To see that $A \backslash B=A \backslash(A \cap B)$ :
$$\begin{array}{rlr} A \backslash(A \cap B) & =A \cap(A \cap B)^c & \ & =A \cap\left(A^c \cup B^c\right) & \text { De Morgan’s law } \ & =\left(A \cap A^c\right) \cup\left(A \cap B^c\right) & \text { distributive law } \ & =\emptyset \cup\left(A \cap B^c\right) & \ & =A \backslash B \end{array}$$
De Morgan’s law distributive law
(b) $A \Delta B=A^c \Delta B^c$ : For any two sets $A$ and $B$
\begin{aligned} & A \backslash B=A \cap B^c \ & =B^c \cap A \quad \text { commutative law } \ & =B^c \cap\left(A^c\right)^c \ & =B^c \backslash A^c \ & \end{aligned}
$\mathrm{So}$
\begin{aligned} A \Delta B & =(A \backslash B) \cup(B \backslash A) \ & =\left(A^c \backslash B^c\right) \cup\left(B^c \backslash A^c\right) \ & =A^c \Delta B^c \end{aligned}
(c) $A \cup B=A \cup(B \backslash A)$ :
\begin{aligned} A \cup B & =A \cup\left((B \cap A) \cup\left(B \cap A^c\right)\right) \quad \text { by part }(\mathrm{d}) \ & =(A \cup(B \cap A)) \cup\left(B \cap A^c\right) \quad \text { associative law } \ & =A \cup\left(B \cap A^c\right) \end{aligned}

(d) $B=(B \cap A) \cup\left(B \cap A^c\right)$ :
\begin{aligned} (B \cap A) \cup\left(B \cap A^c\right) & =B \cap\left(A \cup A^c\right) \quad \text { distributive law } \ & =B \cap S \ & =B \end{aligned}

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