2016 H2 Mathematics Paper 2 Question 5

Probability

Answers

(i)

1142{\frac{11}{42}}

(ii)

411{\frac{4}{11}}

(iii)

41029{\frac{4}{1029}}

Full solutions

(i)

P(win)=P(R ∗)+P(B ∗)+P(Y ∗)=17(36)+27(26)+47(16)=1142  ■\begin{align*} & \textrm{P}\left(\textrm{win}\right) \\ &= \textrm{P}\left(R \, *\right) + \textrm{P}\left(B \, *\right) + \textrm{P}\left(Y \, *\right) \\ &= \frac{1}{7}\left(\frac{3}{6}\right) + \frac{2}{7}\left(\frac{2}{6}\right) + \frac{4}{7}\left(\frac{1}{6}\right) \\ &= \frac{11}{42} \; \blacksquare \end{align*}

(ii)

P(B∣win)=P(B∩win)P(win)=27(26)1142=411  ■\begin{align*} &\textrm{P}\left(B \mid \textrm{win}\right) \\ &=\frac{\textrm{P}\left(B \cap \textrm{win}\right)}{\textrm{P}\left(\textrm{win}\right)} \\ &=\frac{\frac{2}{7}\left(\frac{2}{6}\right)}{\frac{11}{42}} \\ &= \frac{4}{11} \; \blacksquare \end{align*}

(iii)

Required probability
=17(36)×27(26)×47(16)×3!=41029  ■\begin{align*} &= \frac{1}{7}\left(\frac{3}{6}\right) \times \frac{2}{7}\left(\frac{2}{6}\right) \times \frac{4}{7}\left(\frac{1}{6}\right) \times 3! \\ & = \frac{4}{1029} \; \blacksquare \end{align*}