GO Classes DPP | GATE CS, DA | Probability & Statistics | Conditional Probability

Answer 1

can be easily done with tree diagram

Answer 2

By the definition of conditional probability :

$$P(A \mid B \cap C) = \frac{P(A \cap B \cap C)}{P(B \cap C)}$$

Given :

• $P(A) = a$

• $P(B \mid A) = b$

• $P(B \mid A^c) = c$

• $P(C \mid A \cap B) = d$

• $P(C \mid A^c \cap B) = e$

Note: The problem states "$C$ can occur only when $B$ occurs." This means $C$ is a subset of $B$ ($C \subseteq B$), so the event $C$ is exactly the same as the event $B \cap C$. Therefore, finding $P(B \cap C)$ is the same as finding the total probability of $C$ occurring inside $B$.

 

Now, We can use the general multiplication rule (chain rule) of probability to express the joint probability of $A$, $B$, and $C$ occurring together:

$$P(A \cap B \cap C) = P(A) \cdot P(B \mid A) \cdot P(C \mid A \cap B)$$

$$\Rightarrow P(A \cap B \cap C) = a \cdot b \cdot d = abd$$

To find the total probability of $B \cap C$, we can use the Law of Total Probability by partitioning the sample space into $A$ and its complement $A^c$:

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

We already calculated the first part ($abd$). Now let's calculate the second part using the multiplication rule again:

$$P(A^c \cap B \cap C) = P(A^c) \cdot P(B \mid A^c) \cdot P(C \mid A^c \cap B)$$

$$\Rightarrow P(A^c \cap B \cap C) = (1 - a) \cdot c \cdot e$$

$$\Rightarrow P(B \cap C) = P(A \cap B \cap C) + P(A^c \cap B \cap C) = abd + (1 - a)ce$$

Final Probability : 

$$\Rightarrow P(A \mid B \cap C) = \frac{P(A \cap B \cap C)}{P(B \cap C)}$$

$$\boxed{\Rightarrow P(A \mid B \cap C) = \frac{abd}{abd + (1 - a)ce}}$$

Therefore, the correct option is A.