Consider the independent events A and B . Given that ${\rm{P}}(B) = 2{\rm{P}}(A)$ , and ${\rm{P}}(A \cup B) = 0.52$ , find ${\rm{P}}(B)$ .

METHOD 1

for independence ${\rm{P}}(A \cap B) = {\rm{P}}(A) \times {\rm{P}}(B)$     (R1)

expression for ${\rm{P}}(A \cap B)$ , indicating ${\rm{P}}(B) = 2{\rm{P}}(A)$     (A1)

e.g. ${\rm{P}}(A) \times 2{\rm{P}}(A)$ , $x \times 2x$

substituting into ${\rm{P}}(A \cup B) = {\rm{P}}(A) + {\rm{P}}(B) - {\rm{P}}(A \cap B)$     (M1)

correct substitution     A1

e.g. $0.52 = x + 2x - 2{x^2}$ , $0.52 = {\rm{P}}(A) + 2{\rm{P}}(A) - 2{\rm{P}}(A){\rm{P}}(A)$

correct solutions to the equation     (A2)

e.g. $0.2$, $1.3$ (accept the single answer $0.2$)

${\rm{P}}(B) = 0.4$     A1     N6

[7 marks]

METHOD 2

for independence ${\rm{P}}(A \cap B) = {\rm{P}}(A) \times {\rm{P}}(B)$     (R1)

expression for ${\rm{P}}(A \cap B)$ , indicating ${\rm{P}}(A) = \frac{1}{2}{\rm{P}}(B)$     (A1)

e.g. ${\rm{P}}(B) \times \frac{1}{2}{\rm{P}}(B)$ , $x \times \frac{1}{2}x$

substituting into ${\rm{P}}(A \cup B) = {\rm{P}}(A) + {\rm{P}}(B) - {\rm{P}}(A \cap B)$     (M1)

correct substitution     A1

e.g. $0.52 = 0.5x + x - 0.5{x^2}$ , $0.52 = 0.5{\rm{P}}(B) + {\rm{P}}(B) - 0.5{\rm{P}}(B){\rm{P}}(B)$

correct solutions to the equation     (A2)

e.g. 0.4, 2.6 (accept the single answer 0.4)

${\rm{P}}(B) = 0.4$ (accept $x = 0.4$ if x set up as ${\rm{P}}(B)$ )     A1     N6

[7 marks]

Many candidates confused the concept of independence of events with mutual exclusivity, mistakenly trying to use the formula ${\rm{P}}(A \cup B) = {\rm{P}}(A) + {\rm{P}}(B)$ . Those who did recognize that ${\rm{P}}(A \cap B) = {\rm{P}}(A) \times {\rm{P}}(B)$ were often able to find the correct equation, but many were unable to use their GDC to solve it. A few provided two answers without discarding the value greater than one.