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#### karush

##### Well-known member

- Jan 31, 2012

- 3,187

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- Thread starter karush
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- Thread starter
- #1

- Jan 31, 2012

- 3,187

open

Last edited:

- Thread starter
- #3

- Jan 31, 2012

- 3,187

so I have $B\cup A'$

(c) $\displaystyle\frac{n(B\cup A')}{n(U)}=\frac{35}{100}=\frac{7}{20}$ |

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That is correct.so I have $B\cup A'$

(c)

$\displaystyle\frac{n(B\cup A')}{n(U)}=\frac{35}{100}=\frac{7}{20}$