1. Key Concepts and Formulas

• Sample Space ($\Omega$): The set of all possible outcomes of a random experiment.
• Event (A): Any subset of the sample space, $\mathrm{A \subseteq \Omega}$.
• Probability of an Event ($\mathrm{P(A)}$): A real number between 0 and 1 (inclusive) representing the likelihood of event A occurring. It satisfies the following axioms:
  1. $0 \le \mathrm{P(A)} \le 1$ for any event A.
  2. $\mathrm{P(\Omega) = 1}$ (The probability of the certain event is 1).
  3. $\mathrm{P(\phi) = 0}$ (The probability of the impossible event is 0).
  4. For mutually exclusive events A and B ($\mathrm{A \cap B = \phi}$), $\mathrm{P(A \cup B) = P(A) + P(B)}$.
• Complement of an Event ($\overline{A}$): The event that A does not occur. $\overline{A} = \Omega \setminus A$. Its probability is given by $\mathrm{P(\overline{A}) = 1 - P(A)}$.

2. Step-by-Step Solution

The truthfulness of statements (S1) and (S2) depends on the nature of the sample space. In the context of the Joint Entrance Examination (JEE), unless explicitly stated otherwise (e.g., in problems involving geometrical probability), it is almost always assumed that we are dealing with discrete probability spaces. In such spaces, each elementary outcome (individual point in the sample space) is assumed to have a non-zero probability. We will evaluate the statements under this standard JEE assumption.

Step 1: Analyze Statement (S1) - If P(A) = 0, then A = $\phi$

• What we are doing: We are examining whether an event with zero probability must be the empty set (an impossible event) in the standard JEE (discrete) context.
• Why we are doing this: Statement (S1) makes a direct implication from P(A)=0 to A=$\phi$. We need to verify if this implication holds true under the usual assumptions.
• Reasoning (JEE Context - Discrete Probability):
  • Assume, for the sake of contradiction, that A is not empty, but P(A) = 0.
  • If A is not empty, it must contain at least one elementary outcome, say $e$. So, $e \in A$.
  • In discrete probability, unless specified otherwise, every elementary outcome $e$ is considered to have a non-zero probability, i.e., $P(\{e\}) > 0$.
  • Since $A$ contains $e$, and probabilities are non-negative, $P(A) \ge P(\{e\})$.
  • Therefore, if $A$ is non-empty, $P(A)$ must be greater than 0 ($P(A) > 0$).
  • This contradicts our assumption that P(A) = 0.
  • $\text{If A } \neq \phi \implies \text{A contains at least one elementary outcome } e.$
  • $\text{In discrete probability, } P(\{e\}) > 0.$
  • $\text{Since } A \supseteq \{e\}, \text{ it follows that } P(A) \ge P(\{e\}) > 0.$
  • $\text{Thus, if } P(A) = 0, \text{ A must be } \phi.$
• Conclusion for (S1) in JEE Context: Statement (S1) is TRUE under the standard JEE assumption of discrete probability spaces where elementary outcomes have non-zero probabilities.

Step 2: Analyze Statement (S2) - If P(A) = 1, then A = $\Omega$

• What we are doing: We are examining whether an event with probability one must be the entire sample space (a certain event) in the standard JEE (discrete) context.
• Why we are doing this: Statement (S2) makes a direct implication from P(A)=1 to A=$\Omega$. We need to verify if this implication holds true under the usual assumptions.
• Reasoning (JEE Context - Discrete Probability):
  • We can use the complement rule: $\mathrm{P(\overline{A}) = 1 - P(A)}$.
  • Given $\mathrm{P(A) = 1}$, we substitute this into the formula: $\mathrm{P(\overline{A}) = 1 - 1 = 0}$
  • Now, we have $\mathrm{P(\overline{A}) = 0}$. We can apply the conclusion from our analysis of Statement (S1) to the event $\overline{A}$.
  • From Step 1, we established that if the probability of an event is 0 in a discrete sample space, then that event must be the empty set.
  • Therefore, if $\mathrm{P(\overline{A}) = 0}$, it implies that $\overline{A} = \phi$.
  • The complement of A being the empty set ($\phi$) means that A must contain all elements of the sample space $\Omega$. In other words, if nothing is outside A, then A must be everything.
  • $\mathrm{P(A) = 1 \implies P(\overline{A}) = 1 - P(A) = 1 - 1 = 0}$
  • $\text{From (S1) analysis, if } P(\overline{A}) = 0 \implies \overline{A} = \phi$
  • $\text{If } \overline{A} = \phi \implies A = \Omega$
• Conclusion for (S2) in JEE Context: Statement (S2) is TRUE under the standard JEE assumption of discrete probability spaces.

3. Common Mistakes & Tips

• Context is Key: The most common mistake is to confuse the properties of probability in discrete spaces with those in continuous spaces (e.g., geometrical probability). In continuous spaces, an event can have probability 0 without being empty (e.g., the probability of picking a specific point on a line segment is 0, but the point exists), and similarly for probability 1.
• Default JEE Assumption: For JEE problems, always assume a discrete sample space where elementary outcomes have non-zero probabilities unless the problem explicitly describes a continuous scenario (e.g., "a point is chosen uniformly at random from the interval [0,1]").
• Understanding Axioms: A firm grasp of the basic probability axioms and definitions is crucial for such conceptual questions.

4. Summary

In the standard context of JEE Mathematics, which primarily deals with discrete probability spaces where elementary outcomes have non-zero probabilities, both statements (S1) and (S2) are true. If an event has a probability of 0, it must be the empty set (impossible event). Conversely, if an event has a probability of 1, it must be the entire sample space (certain event). This interpretation leads to both statements being correct.

The final answer is $\boxed{\text{both (S1) and (S2) are true}}$, which corresponds to option (A).