If events A and B are independent and $P(A) = \frac{7}{12}$ and $P(A \cap B) = \frac{1}{4}$, find P(B).

Assessment: WAEC SSCE - Further Mathematics - 2008 (Objective) Subject: Further Mathematics

Question 1 Report

$\frac{3}{7}$

$\frac{4}{7}$

$\frac{5}{7}$

$\frac{6}{7}$

Answer Details

If events A and B are independent, then the occurrence of event A does not affect the occurrence of event B, and vice versa. We can use the formula for the probability of the intersection of two independent events: $$P(A \cap B) = P(A) \cdot P(B)$$ Given that $P(A) = \frac{7}{12}$ and $P(A \cap B) = \frac{1}{4}$, we can substitute these values into the formula and solve for $P(B)$: $$\frac{1}{4} = \frac{7}{12} \cdot P(B)$$ Multiplying both sides by $\frac{12}{7}$, we get: $$P(B) = \frac{3}{7}$$ Therefore, the probability of event B is $\frac{3}{7}$. Option (a) is the correct answer.

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If events A and B are independent and \(P(A) = \frac{7}{12}\) and \(P(A \cap B) = \frac{1}{4}\), find P(B).

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If events A and B are independent and \(P(A) = \frac{7}{12}\) and \(P(A \cap B) = \frac{1}{4}\), find P(B).

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