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Question:
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Consider the following statements:
If $A$ and $B$ are independent events, then
$1.$ $A$ and $\bar{B}$ are independent.
$2.$ $\bar{A}$ and $\mathrm{B}$ are independent.
$3.$ $\bar{A}$ and $\bar{B}$ are independent. Which of the above statements is/are correct?
Options:
If $A$ and $B$ are independent events, then
$1.$ $A$ and $\bar{B}$ are independent.
$2.$ $\bar{A}$ and $\mathrm{B}$ are independent.
$3.$ $\bar{A}$ and $\bar{B}$ are independent. Which of the above statements is/are correct?
Solution:
1367 Upvotes
Verified Answer
The correct answer is:
1,2 and 3
Let $\mathrm{A}$ and $\mathrm{B}$ are independent events. $\Rightarrow \quad P(A \cap B)=P(A) \cdot P(B)$
$$
\text { 1. } \begin{aligned}
\text { Consider } \mathrm{P}(\mathrm{A} \cap \overline{\mathrm{B}}) &=\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
&=\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})
\end{aligned}
$$
(from A)
$$
=\mathrm{P}(\mathrm{A})[1-\mathrm{P}(\mathrm{B})]=\mathrm{P}(\mathrm{A}) \mathrm{P}(\overline{\mathrm{B}})
$$
Hence, $\mathrm{A}$ and $\overline{\mathrm{B}}$ are independent.
$2.$ Similarly, $\overline{\mathrm{A}}$ and $\mathrm{B}$ are independent.
$3.$
$$
\begin{aligned}
\text { Consider } \mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}}) &=\mathrm{P}(\overline{\mathrm{A} \cup \mathrm{B}}) \\
&=1-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
=& 1-[\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cup \mathrm{B})] \\
=& 1-\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B}) \\
=&[1-\mathrm{P}(\mathrm{A})][1-\mathrm{P}(\mathrm{B})] \\
=& \mathrm{P}(\overline{\mathrm{A}}) . \mathrm{P}(\overline{\mathrm{B}})
\end{aligned}
$$
Hence, $\overline{\mathrm{A}}$ and $\overline{\mathrm{B}}$ are independent.
$$
\text { 1. } \begin{aligned}
\text { Consider } \mathrm{P}(\mathrm{A} \cap \overline{\mathrm{B}}) &=\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
&=\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})
\end{aligned}
$$
(from A)
$$
=\mathrm{P}(\mathrm{A})[1-\mathrm{P}(\mathrm{B})]=\mathrm{P}(\mathrm{A}) \mathrm{P}(\overline{\mathrm{B}})
$$
Hence, $\mathrm{A}$ and $\overline{\mathrm{B}}$ are independent.
$2.$ Similarly, $\overline{\mathrm{A}}$ and $\mathrm{B}$ are independent.
$3.$
$$
\begin{aligned}
\text { Consider } \mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}}) &=\mathrm{P}(\overline{\mathrm{A} \cup \mathrm{B}}) \\
&=1-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
=& 1-[\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cup \mathrm{B})] \\
=& 1-\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B}) \\
=&[1-\mathrm{P}(\mathrm{A})][1-\mathrm{P}(\mathrm{B})] \\
=& \mathrm{P}(\overline{\mathrm{A}}) . \mathrm{P}(\overline{\mathrm{B}})
\end{aligned}
$$
Hence, $\overline{\mathrm{A}}$ and $\overline{\mathrm{B}}$ are independent.
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