Question

For two given events $A$ and $B, P(A \cap B)$ is

• A. not less than $P(A) + P(B) - 1$
• B. not greater than $P(A) + P(B)$
• C. equal to $P(A) + P(B) - P(A \cup B)$
• D. equal to $P(A) + P(B) + P(A \cup B )$

Answer 1

Answer: (C)

equal to $P(A) + P(B) - P(A \cup B)$

Answer 2

We know that,
$P (A \cap B) = P(A) + P(B) - P(A \cup B)$
Also, $P (A \cup B) \le 1$
$P ( A \cap B)_min$ , when $P ( A \cup B)_max = 1$
$\Rightarrow P ( A \cap B ) \ge P ( A ) + P ( B ) - l$
$\therefore \,$ Option (a) is true.
Again \hspace30mm P(A \cup B) \ge 0
$\therefore \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, P(A \cap B)_{max} , \, when \, \, P(A \cup B)_{min}=0$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, P(A \cap B) \le \, P(A)+P(B)$
$\therefore \, \, \,$ Option (b) is true
Also, $P(A \cap B)=P(A)+P(B)-P(A \cup B.)$ Thus, (c) is
also correct.
Hence, (a), (b), (c) are correct options .