Probability 2 Question 8
8. For two given events $A$ and $B, P(A \cap B)$ is
(1988, 2M)
(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 $P(A)+P(B)+P(A \cup B)$
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Answer:
Correct Answer: 8. (a)
Solution:
- We know that,
$P(A \cap B)=P(A)+P(B)-P(A \cup B)$
Also, $\quad P(A \cup B) \leq 1$
$$ \begin{array}{ll} \therefore & P(A \cap B) _{\min .}, \text { when } P(A \cup B) _{\max }=1 \\ \Rightarrow & P(A \cap B) \geq P(A)+P(B)-1 \end{array} $$
$\therefore$ Option (a) is true.
Again, $\quad P(A \cup B) \geq 0$
$\therefore \quad P(A \cap B) _{\max }$, when $P(A \cup B) _{\min .}=0$
$\Rightarrow \quad P(A \cap B) \leq P(A)+P(B)$
$\therefore$ 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.
