2008 H2 Mathematics Paper 2 Question 7

Probability

Answers

(i)

0.5{0.5}

(ii)

0.512{0.512}

(iii)

764≈0.109{\frac{7}{64}\approx0.109}

Full solutions

(i)

P(A wins second set)=P(AA)+P(BA)=0.6(0.7)+0.4(0.2)=0.5  ■\begin{align*} & \textrm{P}\left(A \textrm{ wins second set}\right) \\ &= \textrm{P}\left(AA\right)+\textrm{P}\left(BA\right) \\ &= 0.6 \left(0.7\right) + 0.4\left(0.2\right) \\ &= 0.5 \; \blacksquare \end{align*}

(ii)

P(A wins match)P(AA)+P(ABA)+P(BAA)=0.6(0.7)+0.6(0.3)(0.2)+0.4(0.2)(0.7)=0.512  ■\begin{align*} & \textrm{P}\left(A \textrm{ wins match}\right) \\ & \textrm{P}\left(AA\right) + \textrm{P}\left(ABA\right) + \textrm{P}\left(BAA\right) \\ &= 0.6 \left(0.7\right) + 0.6\left(0.3\right)\left(0.2\right) + 0.4\left(0.2\right)\left(0.7\right) \\ &= 0.512 \; \blacksquare \end{align*}

(iii)

P(B won first set∣A wins match)=P(B won first set∩A wins match)P(A wins match)=P(BAA)P(A wins match)=0.4(0.2)(0.7)0.512=764  ■\begin{align*} &\textrm{P}\left(B \textrm{ won first set} \mid A \textrm{ wins match}\right) \\ &=\frac{\textrm{P}\left(B \textrm{ won first set} \cap A \textrm{ wins match}\right)}{\textrm{P}\left(A \textrm{ wins match}\right)} \\ &= \frac{\textrm{P}\left(\textrm{BAA}\right)}{\textrm{P}\left(A \textrm{ wins match}\right)} \\ &= \frac{0.4\left(0.2\right)\left(0.7\right)}{0.512} \\ &= \frac{7}{64} \; \blacksquare \end{align*}