Question

Two events $A$ and $B$ are said to be independent, if:

• A. $A$ and $B$ are mutually exclusive.
• B. $P(A'B')=[1–P(A)] [1–P(B)]$
• C. $P(A)=P(B)$
• D. $P(A)+P(B)=1$

Answer 1

Answer: (B)

$P(A'B')=[1–P(A)] [1–P(B)]$

Answer 2

The correct answer is B:$P(A'B')=[1–P(A)] [1–P(B)]$
Two events A and B are said to be independent, if $P(AB) = P(A) × P(B)$
Consider the result given in alternative B.
$P(A'B')=[1–P(A)] [1–P(B)]$
$\implies P(A'\cap B')=1-P(A)-P(B)+P(A).P(B)$
$\implies 1-P(A\cup B)=1-P(A)-P(B)+P(A).P(B)$
$\implies P(A\cup B)=P(A)+P(B)-P(A).P(B)$
$\implies P(A)+P(B)-P(AB)=P(A)+P(B)-P(A).P(B)$
$\implies P(AB)=P(A).P(B)$
This implies that A and B are independent, if $P(A'B')=[1–P(A)] [1–P(B)]$
Distracter Rationale
A. Let $P (A) = m, P (B) = n, 0 < m, n < 1$
$A$ and $B$ are mutually exclusive
$\therefore A\cap B=\phi$
$\implies P(AB)=0$
However, $P(A).P(B)=mn\neq0$
$\therefore P(A).P(B) \neq P(AB)$
C. Let $A$: Event of getting an odd number on throw of a die = {1, 3, 5}
$\implies P(A)=\frac{3}{6}=\frac{1}{2}$
B: Event of getting an even number on throw of a die = {2, 4, 6}
$\implies P(B)=\frac{3}{6}=\frac{1}{2}$
$$Here, $A\cap B=\phi$
$\therefore P(AB)=0$
$P(A).P(B)=\frac{1}{4} \neq0$
$\implies P(A).P(B) \neq P(AB)$
D. From the above example, it can be seen that,
$P(A)+P(B)=\frac{1}{2}+\frac{1}{2}$
However, it cannot be inferred that $A$ and $B$ are independent.
Thus, the correct answer is B.