22M.2.SL.TZ2.4

$\text{P}(A\cup B)=\text{P}\left(A\right)+\text{P}\left(B\right)-\text{P}(A\cap B)=0.68$

substitution of $\text{P}\left(A\right)·\text{P}\left(B\right)$ for $\text{P}(A\cap B)$ in $\text{P}(A\cup B)$         (M1)

$\text{P}\left(A\right)+\text{P}\left(B\right)-\text{P}\left(A\right)\text{P}\left(B\right)\left(=0.68\right)$

substitution of $3\text{P}\left(B\right)$ for $\text{P}\left(A\right)$         (M1)

$3\text{P}\left(B\right)+\text{P}\left(B\right)-3\text{P}\left(B\right)\text{P}\left(B\right)=0.68$  (or equivalent)         (A1)

 

Note: The first two M marks are independent of each other.

 

attempts to solve their quadratic equation         (M1)

$\text{P}\left(B\right)=0.2, 1.133\dots \left(\frac{1}{5}, \frac{17}{15}\right)$

$\text{P}\left(B\right)=0.2 \left(=\frac{1}{5}\right)$          A2

 

Note: Award A1 if both answers are given as final answers for $\text{P}\left(B\right)$.

 

[6 marks]

This question proved difficult for many students. One common error was to use $\text{P}(\text{A}\cup \text{B})=\text{P(A)+P(B)}$, which simplified the problem greatly, resulting in a linear, not a quadratic equation.