Two events A and B will be independent, if

MCQ

Two events $A$ and $B$ will be independent, if

A
$A$ and $B$ are mutually exclusive
✓
$\text{P}(\text{A}\ '\text{B}\ ')=\big[1-\text{P}(\text{A})\big]\big[1-\text{P}(\text{B})\big]$
C
$P(A) = P(B)$
D
$P(A) + P(B) = 1$

✓

Answer

Correct option: B.

$\text{P}(\text{A}\ '\text{B}\ ')=\big[1-\text{P}(\text{A})\big]\big[1-\text{P}(\text{B})\big]$

Two events $A$ and $B$ are said to be independent, if $\text{P}(\text{AB})=\text{P}(\text{A})\times\text{P}(\text{B})$Distracter Rationale.

Let $P(A) = m, P(B) = n, 0 < m, n < 1$

$A$ and $B$ are mutually exclusive.
$\therefore\text{A}\cap\text{B}=\phi$
$\Rightarrow\text{P}(\text{AB})=0$
$\text{However,}\ \text{P}(\text{A})\cdot\text{P}(\text{B})=mn\neq0$
$\therefore\text{P}(\text{A}).\text{P}(\text{B})\neq\text{P}(\text{AB})$

Consider the result given in alternative.

$\text{P}(\text{A}'\text{B}')=\big[1-\text{P}(\text{A})\big]\big[1-\text{P}(\text{B})\big]$
$\Rightarrow\text{P}(\text{A}'\cap\text{B}')=1-\text{P}(\text{A})-\text{P}(\text{B})+\text{P}(\text{A}).\text{P}(\text{B})$
$\Rightarrow1-\text{P}(\text{A}\cup\text{B})=1-\text{P}(\text{A})-\text{P}(\text{B})+\text{P}(\text{A}).\text{P}(\text{B})$
$\Rightarrow\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A})\cdot\text{P}(\text{B})$
$ \Rightarrow\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{AB})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}).\text{P}(\text{B})$
$\Rightarrow\text{P}(\text{AB})=\text{P}(\text{A}).\text{P}(\text{B})$
This implies that $A$ and $B$ are independent, if $\text{P}(\text{A}\ '\text{B}\ ')=\big[1-\text{P}(\text{A})\big]\big[1-\text{P}(\text{B})\big]$

Let $A:$ Event of getting an odd number on throw of a die $= \{1, 3, 5\}$

$\Rightarrow\text{P}(\text{A})=\frac{3}{6}=\frac{1}{2}$
$B:$ Event of getting an even number on throw of a die $= \{2, 4, 6\}$
$\text{P}(\text{B})=\frac{3}{6}=\frac{1}{2}$
Here, $\text{A}\cap\text{B}=\phi$
$\therefore\text{P}(\text{AB})=0 $
$\text{P}(\text{A}).\text{P}(\text{B})=\frac{1}{4}\neq0$
$ \Rightarrow\text{P}(\text{A}).\text{P}(\text{B})\neq\text{P}(\text{AB})$

From the above example, it can be seen that,

$\text{P}(\text{A})+\text{P}(\text{B})=\frac{1}{2}+\frac{1}{2}=1$
However, it cannot be inferred that $A$ and $B$ are independent.
Thus, the correct answer is $B.$

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