2019 Paper 2 Question 6

(a)

(i)

If the formula isn’t known, it can be worked out using a basic formula you must know

$P( X \cup Y ) = P(X) + P(Y) - P( X \cap Y)$

Let $X=A$ and $Y=B \cup C$

$P( A \cup B \cup C) = P(A) + P(B \cup C) - P(A \cap (B \cup C)) \qquad(1)$

Re-use the basic formula

$P( B \cup C ) = P(B) + P(C) - P( B \cap C) \qquad(2)$

Also since

$A \cap (B \cup C) = (A\cap B) \cup (A\cap C)$

then

$P(A \cap (B \cup C)) = P(A\cap B) + P(A\cap C)-P(A\cap B\cap C) \qquad(3)$

Putting (2) and (3) in (1)

$\begin{aligned} P(A \cup B \cup C) &= P(A) + P(B)+P(B) \\ &- P(A \cap B)-P(A \cap C)-P(B \cap C)+P(A\cap B\cap C) \end{aligned}$

(ii)

If $D$ has occurred, then so has $A \cup B \cup C$

$P(A \cup B \cup C|D) =1$

Or, a bit more algebraically

$P(A \cup B \cup C|D) = P( (A \cup B \cup C) \cap D )/P(D)$

Consider the numerator on the RHS

$(A \cup B \cup C) \cap D = (A \cap D) \cup (B \cap D) \cup (C \cap D) = D$

as $D$ is a proper subset of $A \cup B \cup C$ . Thus

$P( (A \cup B \cup C) \cap D ) = P(D)$

leading to the same answer as before.

NST First Year Maths