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.