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If events $B$ and $C$ are dependent on event $A$ and $P(A \hspace{0.1cm}and\hspace{0.1cm} B) = 0.30$, $P(A\hspace{0.1cm} and\hspace{0.1cm} C) = 0.20$ and the dependent events $B$ and $C$ are mutually exclusive and collectively exhaustive, then $P(C/A)$ is equal to ________ ?
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$\color{blue}{P(A\hspace{0.1cm} \cap\hspace{0.1cm} B)  = 0.30}$

$\color{blue}{P(A\hspace{0.1cm} \cap \hspace{0.1cm} C) =  0.20}$

Now, $B$ and $C$ are collectively exhaustive means $B$ & $C$ together exhaust all the possibilities.

∴ The union of  $B$ & $C$ must cover all the events in the entire sample space.

∴ $\color{violet}{P(B \cup C) = 1}$

$B $ & $C$ are also mutually exclusive  means $B $ & $C$ cannot occur simultaneously.

∴ $\color{violet}{P(B \cap C) = 0}$

${P(A\hspace{0.1cm} and\hspace{0.1cm} B) \neq P(A) \times P(B)}\Rightarrow$ because Event $B$ depends on Event $A$.

& ${P(A\hspace{0.1cm} and\hspace{0.1cm} C) \neq P(A) \times P(C)}\Rightarrow$ because Event $C$ depends on Event $A$.

Events A and B are independent events if and only if :

$P(A \cap B) = P(A) \times P(B)$

Otherwise, A and B are called dependent events.


Applying union in both the cases:

$P((A \cap B) \cup (A \cap C)) = 0.3 + 0.2 = 0.5$

As per distributive law

$P(A \cap (B \cup C)) = 0.5$

We already know that, $P(B \cup C) = 1$ as they are collectively exhaustive.

$∴P(A \cap 1) $ or $\color{violet}{P(A) = 0.5} $

$\color{green}{P(C/A) = \dfrac{P(C\cap A)}{P(A)} = \dfrac{0.20}{0.50}} =\color{orange} {0.4}$

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P(A ∩B)=0.3.................................1
P(A ∩C)=0.2.................................2
Since B and C are mutually exclusive and collectively exhaustive=> P(BUC)=1
Now On taking union on both sides of 1 &2
P((A ∩ B) U (A ∩C))=0.3+0.2
P(A ∩(BUC))=0.5
P(A ∩1)=0.5
P(A)=0.5

Now, P(C|A)=P(C∩A)|P(A)=0.2/0.5=0.4
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B and C are mutually exclusive and collectively exhaustive and both the events are dependent on A.

$P(B\cap C)=0, P(B\cup C)=1$

Hence, we can use the law of total probability here [refer https://en.wikipedia.org/wiki/Law_of_total_probability],

$P(A)=P(A\cap B)+P(A\cap C)$

$\implies P(A)=0.3+0.2=0.5$

Now, $P(C|A)=\frac{0.2}{0.5}=0.4$

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