In an LPP, the feasible region represented by the set off constraints 2x + 3y ≤ 18, x + y ≤ 10, x ≥ 0, y ≥ 0 is

Given information:

• $P(A) \neq 0$P(A)=0 (Event A is not impossible)
• $P(B | A) = 1$P(B∣A)=1 (If A happens, then B happens with certainty)

When $P(B | A) = 1$P(B∣A)=1, it means that whenever A occurs, B must also occur.

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The conditional probability formula:

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

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Substituting $P(B | A) = 1$P(B∣A)=1:

$$1 = \frac{P(A \cap B)}{P(A)}$$1=P(A)P(A∩B)​

$$P(A \cap B) = P(A)$$P(A∩B)=P(A)

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The equation $P(A \cap B) = P(A)$P(A∩B)=P(A) means that whenever A happens, both A and B happen together.

In set notation, this means all elements of A are also in B.

Therefore: $A \subset B$A⊂B (A is a subset of B)

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Why other options are incorrect:

$B \subset A$B⊂A: This would mean B is a subset of A, contradicting the established relationship.

$B = \phi$B=ϕ: If B were empty, $P(B | A)$P(B∣A) could not equal 1.

$A = \phi$A=ϕ: This contradicts the given condition $P(A) \neq 0$P(A)=0.

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The answer is: $A \subset B$A⊂B