Data available for profit (₹ Thousands) of a company as | Year | 2004 | 2005 | 2006 | 2007 | 2008 | 2009 | 2010 | |---|---|---|---|---|---|---|---| |…

Statement (A): $P(S | B) = 1$P(S∣B)=1

Using the conditional probability formula:

$$P(S | B) = \frac{P(S \cap B)}{P(B)}$$P(S∣B)=P(B)P(S∩B)​

Since $B \subseteq S$B⊆S, we have $S \cap B = B$S∩B=B

$$P(S | B) = \frac{P(B)}{P(B)} = 1$$P(S∣B)=P(B)P(B)​=1

Statement (A) is TRUE.

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Statement (B): $P(A \cap B) = P(A) + P(B) + P(A \cup B)$P(A∩B)=P(A)+P(B)+P(A∪B)

The addition rule for probability states:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$P(A∪B)=P(A)+P(B)−P(A∩B)

Rearranging:

$$P(A \cap B) = P(A) + P(B) - P(A \cup B)$$P(A∩B)=P(A)+P(B)−P(A∪B)

The given statement has a plus sign instead of a minus sign.

Statement (B) is FALSE.

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Statement (C): $P(\bar{A} | B) = 1 - P(A | B)$P(Aˉ∣B)=1−P(A∣B)

Given that event $B$B has occurred, either $A$A occurs or $\bar{A}$Aˉ occurs:

$$P(A | B) + P(\bar{A} | B) = 1$$P(A∣B)+P(Aˉ∣B)=1

Therefore:

$$P(\bar{A} | B) = 1 - P(A | B)$$P(Aˉ∣B)=1−P(A∣B)

Statement (C) is TRUE.

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Statement (D): $P(A | B) = \frac{P(A \cap B)}{P(B)}, P(B) \neq 0$P(A∣B)=P(B)P(A∩B)​,P(B)=0

This is the definition of conditional probability.

Statement (D) is TRUE.

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Statements (A), (C), and (D) are TRUE.

Statement (B) is FALSE.

The correct answer is: (A), (C) and (D) only.