Question

If $A _ { 1 } , A _ { 2 } , \ldots A _ { n }$ are any n events, then

• A. $P \left( A _ { 1 } \cup A _ { 2 } \cup \ldots \cup A _ { n } \right) = P \left( A _ { 1 } \right) + P \left( A _ { 2 } \right) + \ldots + P \left( A _ { n } \right)$
• B. $P \left( A _ { 1 } \cup A _ { 2 } \cup \ldots \cup A _ { n } \right) > P \left( A _ { 1 } \right) + P \left( A _ { 2 } \right) + \ldots + P \left( A _ { n } \right)$
• C. $P \left( A _ { 1 } \cup A _ { 2 } \cup \ldots \cup A _ { n } \right) \leq P \left( A _ { 1 } \right) + P \left( A _ { 2 } \right) + \ldots + P \left( A _ { n } \right)$
• D. None of these

Answer 1

Answer: (C)

$P \left( A _ { 1 } \cup A _ { 2 } \cup \ldots \cup A _ { n } \right) \leq P \left( A _ { 1 } \right) + P \left( A _ { 2 } \right) + \ldots + P \left( A _ { n } \right)$

Answer 2

For any two events $A$ and we have

$P ( A \cup B ) = P ( A ) + P ( B ) - P ( A \cap B )$

$\therefore P ( A \cup B ) \leq P ( A ) + P ( B )$

Using principle of mathematical induction, it can be easily established that $P \left( \bigcup _ { i = 1 } ^ { n } A _ { i } \right) \leq \sum _ { i = 1 } ^ { n } P \left( A _ { i } \right)$.