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, \(\displaystyle{S}_{{{1}}}\cup{S}_{{{2}}}={\left\lbrace{32},{23},{29},{31},{37},{41},{43},{47},{53},{59}\right\rbrace}\)

\(\displaystyle{S}_{{{1}}}\cap{S}_{{{2}}}\cap{S}_{{{3}}}=\phi\)

\(\displaystyle{I}.{P}{\left({S}_{{{3}}}\right)}={\frac{{{14}}}{{{40}}}}={\frac{{{7}}}{{{20}}}}\)

\(\displaystyle{I}{I}.{P}{\left({S}_{{{1}}}\cup{S}_{{{2}}}\right)}={\frac{{{10}}}{{{40}}}}={\frac{{{1}}}{{{4}}}}\)

\(\displaystyle{I}{I}{I}.{P}{\left({S}_{{{1}}}\cap{S}_{{{2}}}\cap{S}_{{{3}}}\right)}={\frac{{{0}}}{{{40}}}}={0}\)

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, \(\displaystyle{S}_{{{1}}}\cup{S}_{{{2}}}={\left\lbrace{32},{23},{29},{31},{37},{41},{43},{47},{53},{59}\right\rbrace}\)

\(\displaystyle{S}_{{{1}}}\cap{S}_{{{2}}}\cap{S}_{{{3}}}=\phi\)

\(\displaystyle{I}.{P}{\left({S}_{{{3}}}\right)}={\frac{{{14}}}{{{40}}}}={\frac{{{7}}}{{{20}}}}\)

\(\displaystyle{I}{I}.{P}{\left({S}_{{{1}}}\cup{S}_{{{2}}}\right)}={\frac{{{10}}}{{{40}}}}={\frac{{{1}}}{{{4}}}}\)

\(\displaystyle{I}{I}{I}.{P}{\left({S}_{{{1}}}\cap{S}_{{{2}}}\cap{S}_{{{3}}}\right)}={\frac{{{0}}}{{{40}}}}={0}\)