Question

If A and B are independent events, then which of the following statements are correct?
[a] A and B are mutually exclusive
[b] A and B’ are independent events
[c] A’ and B’ are independent events
[d] $\text{P}\left( \text{A/B} \right)+\text{P}\left( \text{A }\\!\\!'\\!\\!\text{ /B} \right)=1$

Answer 1

Hint : Two events are said to be independent if the probability of occurrence of one event is not affected by the occurrence or non-occurrence of the other event, i.e. if A and B are independent events then P(A/B) = P(A). Using this definition of the independent events verify which of the above-given statements are always true.

Complete step by step solution :
We have A and B are independent events,
Hence P(A/B) = P(A)
Hence $\dfrac{\text{P}\left( \text{A}\bigcap \text{B} \right)}{\text{P}\left( \text{B} \right)}=\text{P}\left( \text{A} \right)$
Multiplying both sides by P(B), we get
$\text{P}\left( \text{A}\bigcap \text{B} \right)=\text{P}\left( \text{A} \right)\text{P}\left( \text{B} \right)$
Two events A and B are said to be mutually exclusive if the sets A and B are disjoint.
Consider two events A and B such that A has non-zero probability and B = S.
Since $\text{A}\subset \text{S}$, we have $\text{A}\bigcap \text{B=A}\ne \phi$. Hence the events are not mutually exclusive.
However, $\text{P}\left( \text{A}\bigcap \text{B} \right)=\text{P}\left( \text{A} \right)$ and $\text{P}\left( \text{A} \right)\text{P}\left( \text{B} \right)=\text{P}\left( \text{A} \right)$.
Hence the events are independent but not mutually exclusive. Hence option [a] is incorrect.
Now if A and B are independent then, $\text{P}\left( \text{A}\bigcap \text{B} \right)=\text{P}\left( \text{A} \right)\text{P}\left( \text{B} \right)$
Now we know that $\text{A=A}\bigcap \left( \text{B}\bigcup \text{B }\\!\\!'\\!\\!\text{ } \right)=\left( \text{A}\bigcap \text{B} \right)\bigcup \left( \text{A}\bigcap \text{B }\\!\\!'\\!\\!\text{ } \right)$
Hence $\text{P}\left( \text{A} \right)=\text{P}\left( \left( \text{A}\bigcap \text{B} \right)\bigcup \left( \text{A}\bigcap \text{B }\\!\\!'\\!\\!\text{ } \right) \right)$
Since the events $\left( \text{A}\bigcap \text{B} \right)$ and $\left( \text{A}\bigcap \text{B }\\!\\!'\\!\\!\text{ } \right)$ are disjoint, these events are mutually exclusive
Hence $\text{P}\left( \text{A} \right)\text{=P}\left( \left( \text{A}\bigcap \text{B} \right)\bigcup \left( \text{A}\bigcap \text{B }\\!\\!'\\!\\!\text{ } \right) \right)=\text{P}\left( \text{A}\bigcap \text{B} \right)+\text{P}\left( \text{A}\bigcap \text{B }\\!\\!'\\!\\!\text{ } \right)$
Hence, we have
$\begin{aligned} & \text{P}\left( \text{A} \right)=\text{P}\left( \text{A} \right)\text{P}\left( \text{B} \right)+\text{P}\left( \text{A}\bigcap \text{B }\\!\\!'\\!\\!\text{ } \right) \\\ & \Rightarrow \text{P}\left( \text{A}\bigcap \text{B }\\!\\!'\\!\\!\text{ } \right)=\text{P}\left( \text{A} \right)\left( 1-\text{P}\left( \text{B} \right) \right) \\\ \end{aligned}$
We know that P(B’) = 1-P(B)
Hence we have $\text{P}\left( \text{A}\bigcap \text{B }\\!\\!'\\!\\!\text{ } \right)=\text{P}\left( \text{A} \right)\text{P}\left( \text{B }\\!\\!'\\!\\!\text{ } \right)$
Hence the events A and B’ are independent,
Since A and B’ are independent, we have A’ and B’ are also independent.
Hence options [b] and [c] are correct.
Now we know that $\text{P}\left( \text{A/B} \right)=\dfrac{\text{P}\left( \text{A}\bigcap \text{B} \right)}{\text{P}\left( \text{B} \right)}$
Hence, we have
$\text{P}\left( \text{A/B} \right)+\text{P}\left( \text{A }\\!\\!'\\!\\!\text{ /B} \right)=\dfrac{\text{P}\left( \text{A}\bigcap \text{B} \right)}{\text{P}\left( \text{B} \right)}+\dfrac{\text{P}\left( \text{A }\\!\\!'\\!\\!\text{ }\bigcap \text{B} \right)}{\text{P}\left( \text{B} \right)}=\dfrac{\text{P}\left( \text{A}\bigcap \text{B} \right)+\text{P}\left( \text{A }\\!\\!'\\!\\!\text{ }\bigcap \text{B} \right)}{\text{P}\left( \text{B} \right)}$
Since $\text{A}\bigcap \text{B}$ and $\text{A }\\!\\!'\\!\\!\text{ }\bigcap \text{B}$ are mutually exclusive events, we have
$\text{P}\left( \text{A}\bigcap \text{B} \right)+\text{P}\left( \text{A }\\!\\!'\\!\\!\text{ }\bigcap \text{B} \right)=\text{P}\left( \left( \text{A}\bigcap \text{B} \right)\bigcup \left( \text{A }\\!\\!'\\!\\!\text{ }\bigcap \text{B} \right) \right)=\text{P}\left( \left( \text{A}\bigcup \text{A }\\!\\!'\\!\\!\text{ } \right)\bigcap \text{B} \right)=\text{P}\left( \text{B} \right)$
Hence, we have
$\text{P}\left( \text{A/B} \right)+\text{P}\left( \text{A }\\!\\!'\\!\\!\text{ /B} \right)=\dfrac{\text{P}\left( \text{B} \right)}{\text{P}\left( \text{B} \right)}=1$
Hence option[d] is correct.
Hence options [b] , [c] and [d] are correct.

Note : [1] Remember the above results. These results are often used in probability.
[2] A rather easier way to prove the above statement is as follows:
A and B’ will be independent if
$P\left( B'/A \right)=P\left( B' \right)$
We know that if $A\subset B$, then $P\left( B-A \right)=P\left( B \right)-P\left( A \right)$
Now, we have
$P\left( B'/A \right)=\dfrac{P\left( B'\bigcap A \right)}{P\left( A \right)}=\dfrac{P\left( \left( S-B \right)\bigcap A \right)}{P\left( A \right)}=\dfrac{P\left( S\bigcap A-B\bigcap A \right)}{P\left( A \right)}$
Now since $B\bigcap A\subset S\bigcap A$, we have
$P\left( B'/A \right)=\dfrac{P\left( S\bigcap A \right)-P\left( B\bigcap A \right)}{P\left( A \right)}=\dfrac{P\left( A \right)-P\left( A \right)P\left( B \right)}{P\left( A \right)}=1-P\left( B \right)=P\left( B' \right)$
Hence A and B’ are independent events.