Answer
a. 300
b. 150
c. 175
d. 100
Work Step by Step
$|A_1|=100$, $|A_2|=100$, $|A_3|=100$
a. We use the Principle of Inclusion-Exclusion for three sets, $|A_1 \cup A_2 \cup A_3|=|A_1|+|A_2|+|A_3|-|A_1\cap A_2|-|A_1\cap A_3|-|A_2\cap A_3|+|A_1\cap A_2\cap A_3|$. The sets are disjoint means $|A_1\cap A_2|$, $|A_1\cap A_3|$, $|A_2\cap A_3|$, $|A_1\cap A_2\cap A_3|$ are all 0. Then $|A_1 \cup A_2 \cup A_3|=|A_1|+|A_2|+|A_3|-|A_1\cap A_2|-|A_1\cap A_3|-|A_2\cap A_3|+|A_1\cap A_2\cap A_3|=100+100+100-0-0-0+0=300$.
b. $|A_1\cap A_2|=50$, $|A_1\cap A_3|=50$, $|A_2\cap A_3|=50$, $|A_1\cap A_2\cap A_3|=0$. Then $|A_1 \cup A_2 \cup A_3|=|A_1|+|A_2|+|A_3|-|A_1\cap A_2|-|A_1\cap A_3|-|A_2\cap A_3|+|A_1\cap A_2\cap A_3|=100+100+100-50-50-50+0=150$.
c. $|A_1\cap A_2|=50$, $|A_1\cap A_3|=50$, $|A_2\cap A_3|=50$, $|A_1\cap A_2\cap A_3|=25$. Then $|A_1 \cup A_2 \cup A_3|=|A_1|+|A_2|+|A_3|-|A_1\cap A_2|-|A_1\cap A_3|-|A_2\cap A_3|+|A_1\cap A_2\cap A_3|=100+100+100-50-50-50+25=175$.
d. The sets are equal. This means that $|A_1\cap A_2|$, $|A_1\cap A_3|$, $|A_2\cap A_3|$, and $|A_1\cap A_2\cap A_3|$ are all equal to $100$. Then $|A_1 \cup A_2 \cup A_3|=|A_1|+|A_2|+|A_3|-|A_1\cap A_2|-|A_1\cap A_3|-|A_2\cap A_3|+|A_1\cap A_2\cap A_3|=100+100+100-100-100-100+100=100$.
