a) State in words the relationship between two events R and S when P(R ∩ S) = 0 - Edexcel - A-Level Maths Statistics - Question 2 - 2012 - Paper 1

To find P(B), we can use the addition rule for probabilities. Given that P(A ∪ B) = \frac{2}{3} and P(A) = \frac{1}{4}, we can write the equation:

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

Since A and B are independent, we know that:

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

Substituting in the values:

$\frac{2}{3} = \frac{1}{4} + P(B) - \frac{1}{4} \cdot P(B)$

Now, rearranging gives:

$\frac{2}{3} = \frac{1}{4} + P(B)(1 - \frac{1}{4})$

This simplifies to:

$\frac{2}{3} - \frac{1}{4} = \frac{3}{4} P(B)$

Finding a common denominator (12) for the left side:

$\frac{8}{12} - \frac{3}{12} = \frac{3}{4} P(B)$

This results in:

$\frac{5}{12} = \frac{3}{4} P(B)$

Finally, solving for P(B):

$P(B) = \frac{5}{12} \cdot \frac{4}{3} = \frac{5}{9}.$

Thus, P(B) = \frac{5}{9}.

Using the independence of events A and B, we can compute:

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

From previous calculations, we know that:

• P(A) = \frac{1}{4}
• P(B) = \frac{5}{9}

Thus:

$P(A \cap B) = \frac{1}{4} \cdot \frac{5}{9} = \frac{5}{36}.$

So, P(A ∩ B) = \frac{5}{36}.

To find P(B | A), we use the formula for conditional probability:

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

From our previous calculations, we have:

• P(A ∩ B) = \frac{5}{36}
• P(A) = \frac{1}{4}

Thus:

$P(B | A) = \frac{\frac{5}{36}}{\frac{1}{4}} = \frac{5}{36} \cdot \frac{4}{1} = \frac{20}{36} = \frac{5}{9}.$

So, P(B | A) = \frac{5}{9}.