Question:
Let \( E_1, E_2 \) and \( E_3 \) be mutually independent events.
Statement I: \( E_1 \) and \( E_2 \cup E_3 \) are independent.
Statement II: \( E_1 \) and \( E_2 \cap E_3 \) are independent.
Which one of the following options is correct?
Let \( E_1, E_2 \) and \( E_3 \) be mutually independent events.
Statement I: \( E_1 \) and \( E_2 \cup E_3 \) are independent.
Statement II: \( E_1 \) and \( E_2 \cap E_3 \) are independent.
Which one of the following options is correct?
Show Hint
For mutually independent events:
\[
P(A\cap B)=P(A)P(B)
\]
and
\[
P(A\cap B\cap C)=P(A)P(B)P(C)
\]
These properties can often be extended to unions and intersections through algebraic manipulation of probabilities.
Updated On: Jun 17, 2026
- Both I and II are true
- Only I is true
- Only II is true
- Both I and II are false
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Concept:
If events are mutually independent, then:
\[
P(E_i\cap E_j)=P(E_i)P(E_j)
\]
and
\[
P(E_1\cap E_2\cap E_3)
=
P(E_1)P(E_2)P(E_3)
\]
We verify both statements individually.
Step 1: Check Statement I.
We need to test whether: \[ E_1 \text{ and } (E_2\cup E_3) \] are independent. We calculate: \[ P(E_1\cap(E_2\cup E_3)) \] Using distributive law: \[ = P((E_1\cap E_2)\cup(E_1\cap E_3)) \] Applying addition theorem: \[ = P(E_1\cap E_2)+P(E_1\cap E_3)-P(E_1\cap E_2\cap E_3) \] Using mutual independence: \[ = P(E_1)P(E_2)+P(E_1)P(E_3)-P(E_1)P(E_2)P(E_3) \] Factorizing: \[ = P(E_1)\left[P(E_2)+P(E_3)-P(E_2)P(E_3)\right] \] But, \[ P(E_2\cup E_3) = P(E_2)+P(E_3)-P(E_2)P(E_3) \] Hence, \[ P(E_1\cap(E_2\cup E_3)) = P(E_1)P(E_2\cup E_3) \] Therefore, \[ E_1 \text{ and } (E_2\cup E_3) \] are independent. Thus Statement I is TRUE.
Step 2: Check Statement II.
We need to verify whether: \[ E_1 \text{ and } (E_2\cap E_3) \] are independent. Now, \[ P(E_1\cap(E_2\cap E_3)) = P(E_1\cap E_2\cap E_3) \] Using mutual independence: \[ = P(E_1)P(E_2)P(E_3) \] Also, \[ P(E_2\cap E_3)=P(E_2)P(E_3) \] Therefore, \[ P(E_1)P(E_2\cap E_3) = P(E_1)P(E_2)P(E_3) \] Hence, \[ P(E_1\cap(E_2\cap E_3)) = P(E_1)P(E_2\cap E_3) \] Therefore, \[ E_1 \text{ and } (E_2\cap E_3) \] are independent. Thus Statement II is also TRUE. Hence both statements are correct. \[ \boxed{\text{Both I and II are true}} \]
Step 1: Check Statement I.
We need to test whether: \[ E_1 \text{ and } (E_2\cup E_3) \] are independent. We calculate: \[ P(E_1\cap(E_2\cup E_3)) \] Using distributive law: \[ = P((E_1\cap E_2)\cup(E_1\cap E_3)) \] Applying addition theorem: \[ = P(E_1\cap E_2)+P(E_1\cap E_3)-P(E_1\cap E_2\cap E_3) \] Using mutual independence: \[ = P(E_1)P(E_2)+P(E_1)P(E_3)-P(E_1)P(E_2)P(E_3) \] Factorizing: \[ = P(E_1)\left[P(E_2)+P(E_3)-P(E_2)P(E_3)\right] \] But, \[ P(E_2\cup E_3) = P(E_2)+P(E_3)-P(E_2)P(E_3) \] Hence, \[ P(E_1\cap(E_2\cup E_3)) = P(E_1)P(E_2\cup E_3) \] Therefore, \[ E_1 \text{ and } (E_2\cup E_3) \] are independent. Thus Statement I is TRUE.
Step 2: Check Statement II.
We need to verify whether: \[ E_1 \text{ and } (E_2\cap E_3) \] are independent. Now, \[ P(E_1\cap(E_2\cap E_3)) = P(E_1\cap E_2\cap E_3) \] Using mutual independence: \[ = P(E_1)P(E_2)P(E_3) \] Also, \[ P(E_2\cap E_3)=P(E_2)P(E_3) \] Therefore, \[ P(E_1)P(E_2\cap E_3) = P(E_1)P(E_2)P(E_3) \] Hence, \[ P(E_1\cap(E_2\cap E_3)) = P(E_1)P(E_2\cap E_3) \] Therefore, \[ E_1 \text{ and } (E_2\cap E_3) \] are independent. Thus Statement II is also TRUE. Hence both statements are correct. \[ \boxed{\text{Both I and II are true}} \]
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