2019 H2 Mathematics Paper 2 Question 8

Probability

Answers

(ia)
2356{\frac{23}{56}}
(ib)
1328{\frac{13}{28}}
(iia)
155{\frac{1}{55}}
(iib)
21110{\frac{21}{110}}

(iii)

  • Orange Bird, Yellow Bird
  • Orange Bird, White Rider
  • White Horse, Yellow Bird
  • White Horse, White Rider

Full solutions

(ia)
P(Horse or Rider)=2356  ■\begin{align*} & \textrm{P}\left(\textrm{Horse or Rider}\right) \\ &= \frac{23}{56} \; \blacksquare \end{align*}
(ib)
P(Dog or Bird but not White)=2656=1328  ■\begin{align*} & \textrm{P}\left(\textrm{Dog or Bird but not White}\right) \\ &= \frac{26}{56} \\ &= \frac{13}{28} \; \blacksquare \end{align*}
(iia)
P(both Horses, neither Orange)=856×755=155  ■\begin{align*} & \textrm{P}\left(\textrm{both Horses, neither Orange}\right) \\ & = \frac{8}{56} \times \frac{7}{55} \\ & = \frac{1}{55} \; \blacksquare \end{align*}
(iib)
Case 1: one Yellow Dog, one non-Yellow non-Dog
Probability
=756×3255×2!=855\begin{align*} &= \frac{7}{56} \times \frac{32}{55} \times 2! \\ &= \frac{8}{55} \\ \end{align*}
Case 2: one non-Yellow Dog, one Yellow non-Dog
Probability
=1056×755×2!=122\begin{align*} &= \frac{10}{56} \times \frac{7}{55} \times 2! \\ &= \frac{1}{22} \\ \end{align*}
Required probability
=855+122=21110  ■\begin{align*} &= \frac{8}{55} + \frac{1}{22} \\ &= \frac{21}{110} \; \blacksquare \end{align*}

(iii)

Let x{x} and y{y} denote the number of Gerri's two favourite characters
P(two favourites)=177x56×y55×2!=177xy=20\begin{align*} \textrm{P}\left(\textrm{two favourites}\right) &= \frac{1}{77} \\ \frac{x}{56} \times \frac{y}{55} \times 2! &= \frac{1}{77} \\ xy &= 20 \\ \end{align*}
Hence the possibilities are
  • Orange Bird and Yellow Bird ■{\blacksquare}
  • Orange Bird and White Rider ■{\blacksquare}
  • White Horse and Yellow Bird ■{\blacksquare}
  • White Horse and White Rider ■{\blacksquare}