The Universal Set, U, consists of the natural numbers from 20 to 60 incluive Define or describe in words the following three (3) sets: factors of 64, prime numbers, and multiples of 3.

The Universal Set, U, consists of the natural numbers from 20 to 60 incluive
a. Define or describe in words the following three (3) sets: factors of 64, prime numbers, and multiples of 3.
b. List the elements in each of your sets:
$$\displaystyle{A}={\lbrace}$$
$$\displaystyle{B}={\lbrace}$$
$$\displaystyle{C}={\lbrace}$$
c. Determine the probability of each of the following:
$$\displaystyle{I}.{P}{\left({C}\right)}$$
$$\displaystyle{I}{I}.{P}{\left({A}\cup{B}\right)}$$
$$\displaystyle{I}{I}{I}.{P}{\left({A}\cap{B}\cap{C}\right)}$$
$$\displaystyle{I}{V}.{P}{\left({B}{C}\right)}$$
$$\displaystyle{V}.{P}{\left({\left({A}{B}\right)}\cap{C}\right)}$$

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Step 1: Given,
The universal set U, consist of the natural numbers from 20 to 60 inclusive. We have to answer the following...
Step 2: Explanation
Solution (a).
If we have 3 sets,
$$\displaystyle{S}_{{{1}}}=$$ factors of 64 i.e. 1,2,4,8,16,32,64
$$\displaystyle{S}_{{{2}}}=$$ prime numbers
$$\displaystyle{S}_{{{3}}}=$$ multiple of 3
Now describe the sets in words,
$$\displaystyle{S}_{{{1}}}={\left\lbrace{x}:{x}\right.}$$ is a factor of 64 and $$\displaystyle{20}\leq{x}\leq{60}\rbrace$$
$$\displaystyle{S}_{{{2}}}={\left\lbrace{x}:{x}\right.}$$ is prime and $$\displaystyle{20}\leq{x}\leq{60}\rbrace$$
$$\displaystyle{S}_{{{3}}}={\left\lbrace{x}:{x}\right.}$$ is multiple of 3 and $$\displaystyle{20}\leq{x}\leq{60}\rbrace$$
Solution (b).
List of the elements
$$\displaystyle{S}_{{{1}}}={\left\lbrace{32}\right\rbrace}$$
$$\displaystyle{S}_{{{2}}}={\left\lbrace{23},{29},{31},{37},{41},{43},{47},{53},{59}\right\rbrace}$$
$$\displaystyle{S}_{{{3}}}={\left\lbrace{21},{24},{27},{30},\ldots,{57},{60}\right\rbrace}$$
Solution (c).
Determine the probability
Since we have from the b part, $$\displaystyle{S}_{{{1}}}\cup{S}_{{{2}}}={\left\lbrace{32},{23},{29},{31},{37},{41},{43},{47},{53},{59}\right\rbrace}$$
$$\displaystyle{S}_{{{1}}}\cap{S}_{{{2}}}\cap{S}_{{{3}}}=\phi$$
$$\displaystyle{I}.{P}{\left({S}_{{{3}}}\right)}={\frac{{{14}}}{{{40}}}}={\frac{{{7}}}{{{20}}}}$$
$$\displaystyle{I}{I}.{P}{\left({S}_{{{1}}}\cup{S}_{{{2}}}\right)}={\frac{{{10}}}{{{40}}}}={\frac{{{1}}}{{{4}}}}$$
$$\displaystyle{I}{I}{I}.{P}{\left({S}_{{{1}}}\cap{S}_{{{2}}}\cap{S}_{{{3}}}\right)}={\frac{{{0}}}{{{40}}}}={0}$$