Question

If $A$ and $B$ are any two events associated with random experiment, then by the definition of probability, which of following is true?
(i) $P(A \cup B)=\Sigma P\left(\omega_i\right) \forall \omega_i \in A \cup B$
(ii) $P(A)+P(B)=P(A \cup B)+P(A \cap B)$

• A. Only (i)
• B. Only (ii)
• C. Both (i) and (ii)
• D. None of these

Answer 1

Answer: (C)

In general, if $A$ and $B$ are any two events associated with a random experiment, then by the definition of probability of an event, we have
$P(A \cup B)=\Sigma P\left(\omega_i\right), \forall \omega_i \in A \cup B$
Since, $A \cup B=(A-B) \cup(A \cap B) \cup(B-A)$,
We have
$P(A \cup B) $
$=\left[\Sigma P\left(\omega_i\right) \forall \omega_i \in(A-B)\right]+\left[\Sigma P\left(\omega_i\right)\forall \omega_i \in A \cap B\right]$
$ +\left[\Sigma P\left(\omega_i\right) \forall \omega_i \in B-A\right]$
(because $A-B, A \cap B$ and $B-A$ are mutually exclusive) .....(i)
Also, $P(A)+P(B)=\left[\Sigma p\left(\omega_i\right) \forall \omega_i \in A\right]+\left[\Sigma P\left(\omega_i\right) \forall \omega_i \in B\right]$
$=\left[\Sigma P\left(\omega_i\right) \forall \omega_i \in(A-B) \cup(A \cap B)\right]+$
$\left[\Sigma P\left(\omega_i\right) \forall \omega_i \in(B-A) \cup(A \cap B)\right]$
$=\left[\Sigma P\left(\omega_i\right) \forall \omega_i \in(A-B)\right]+\left[\Sigma P\left(\omega_i\right) \forall \omega_i \in(A \cap B)\right]+$
$\left[\Sigma P\left(\omega_t\right) \forall \omega_l \in(B-A)\right]+\left[\Sigma P\left(\omega_l\right) \forall \omega_l \in(A \cap B)\right]$
$=P(A \cup B)+\left[\Sigma P\left(\omega_i\right) \forall \omega_i \in A \cap B\right] $ [using Eq. (i) $]$
$=P(A \cup B)+P(A \cap B)$
Hence, $P(A)+P(B)=P(A \cup B)+P(A \cap B)$