2.2.1 problem 33

(a)

$\frac{1}{2^{\left|C\right|}}$

(b)

For each element of $C$ we have four options for whether this element is in $A$ and/or $B$ and each option has equal probability of occurring. An element $x\in C$ is in $A\subseteq B$ if and only if ($x\in B\text{and}x\in A$) or ($x\in B\text{and}x\notin A$) or ($x\notin B\text{and}\notin A$), i.e., 3 times out of 4. Thus, the probability that $x\in A\subseteq B$ is $3∕4$ and $P(A\subseteq B)={(3∕4)}^{100}$.

Another way to see this is by using the naive definition of probability. The sample space consists of 100 binary pairs where 1 in the 1st slot of the $i$-th pair indicates that the $i$-th element of $C$ is in $A$ and 1 in the 2nd slot indicates that the element is in $B$. Hence, $\left|S\right|=4^{100}$. The number of elements in the set $X$ of outcomes corresponding to $A\subseteq B$ can be counted as

$$\left|X\right|=\sum _{i=0}^{100}\left(\frac{100}{i}\right)2^i=3^{100}.$$

The binomial coefficient accounts for the number of $i$-element subsets $B$ of $C$ and $2^i$ is the number of all subsets $A$ of $B$. This gives $P(A\subseteq B)=\left|X\right|∕\left|S\right|={(3∕4)}^{100}.$

(c)

Let $p$ be a randomly selected person from $C$ sampled without replacement.

$P(p\in A\cup p\in B)=\frac{1}{2}+\frac{1}{2}-\frac{1}{4}=\frac{3}{4}$.

$P(A\cup B=C)={\left(P\right(p\in A\cup p\in B\left)\right)}^{\left|C\right|}={\left(\frac{3}{4}\right)}^{\left|C\right|}.$