2012 H2 Mathematics Paper 2 Question 7

Permutations and Combinations (P&C)

Answers

(i)

215{\frac{2}{15}}

(ii)

3435{\frac{34}{35}}

(iii)

2455{\frac{2}{455}}

(iv)

43273{\frac{43}{273}}

(v)

17{\frac{1}{7}}

Full solutions

(i)

Let S{S} denote the event that the sisters are next to each other
P(S)=14!×2!15!=215  ■\begin{align*} \textrm{P}\left(S\right) &= \frac{14! \times 2!}{15!} \\ &= \frac{2}{15} \; \blacksquare \end{align*}

(ii)

Let B{B} denote the event that all the brothers are next to one another
P(B′)=1−P(B)=13!×3!15!=3435  ■\begin{align*} \textrm{P}\left(B'\right) &= 1 - \textrm{P}\left(B\right) \\ &= \frac{13! \times 3!}{15!} \\ &= \frac{34}{35} \; \blacksquare \end{align*}

(iii)

P(S∩B)=12!×3!×2!15!=2455  ■\begin{align*} & \textrm{P}\left(S \cap B\right) \\ &= \frac{12! \times 3! \times 2!}{15!} \\ &= \frac{2}{455} \; \blacksquare \end{align*}

(iv)

P(S∪B)=P(S)+P(B)−P(S∩B)=215+135−2455=43273  ■\begin{align*} & \textrm{P}\left(S \cup B\right) \\ & = \textrm{P}\left(S\right) + \textrm{P}\left(B\right) - \textrm{P}\left(S \cap B\right) \\ &= \frac{2}{15} + \frac{1}{35} - \frac{2}{455} \\ &= \frac{43}{273} \; \blacksquare \end{align*}

(v)

Probability required
=13!×2!14!=17  ■\begin{align*} &= \frac{13! \times 2!}{14!} \\ &= \frac{1}{7} \; \blacksquare \end{align*}