2015 H2 Mathematics Paper 2 Question 9

Probability

Answers

(i)

0.4{0.4}

(ii)

0.185{0.185}

(iii)

Greatest P(A′∩B′∩C′)=0.33{\textrm{P}\left(A' \cap B' \cap C'\right) = 0.33}
Least P(A′∩B′∩C′)=0.165{\textrm{P}\left(A' \cap B' \cap C'\right) = 0.165}

Full solutions

(i)

Since A{A} and B{B} are independent,
P(B∣A)=P(B)=0.4  ■\begin{align*} \textrm{P}\left(B \mid A\right) &= \textrm{P}\left(B\right) \\ &= 0.4 \; \blacksquare \end{align*}

(ii)

Since A{A} and B{B} are independent,
P(A∩B)=P(A)⋅P(B)=0.45×0.4=0.18\begin{align*} \textrm{P}\left(A \cap B\right) &= \textrm{P}\left(A\right) \cdot \textrm{P}\left(B\right) \\ &= 0.45 \times 0.4 \\ &= 0.18 \end{align*}
Since A{A} and C{C} are independent,
P(A∩C)=P(A)⋅P(C)=0.45×0.3=0.135\begin{align*} \textrm{P}\left(A \cap C\right) &= \textrm{P}\left(A\right) \cdot \textrm{P}\left(C\right) \\ &= 0.45 \times 0.3 \\ &= 0.135 \end{align*}
Given B{B} and C{C} are independent,
P(B∩C)=P(B)⋅P(C)=0.4×0.3=0.12\begin{align*} \textrm{P}\left(B \cap C\right) &= \textrm{P}\left(B\right) \cdot \textrm{P}\left(C\right) \\ &= 0.4 \times 0.3 \\ &= 0.12 \end{align*}
From the Venn diagram,
P(A′∩B′∩C′)=0.185  ■\textrm{P}\left(A' \cap B' \cap C'\right) = 0.185 \; \blacksquare

(iii)

Let x{x} denote P(B∩C∩A′){\textrm{P}\left(B \cap C \cap A'\right)}
From the Venn diagram, since all probabilities are between 0{0} and 1,{1,}
0≤x≤0.1650 \leq x \leq 0.165
Since P(A′∩B′∩C′)=0.165+x,\textrm{Since } \textrm{P}\left(A' \cap B' \cap C'\right) = 0.165+x,
0.165≤P(A′∩B′∩C′)≤0.330.165 \leq \textrm{P}\left(A' \cap B' \cap C'\right) \leq 0.33
Greatest possible P(A′∩B′∩C′)=0.33  ■Least possible P(A′∩B′∩C′)=0.165  ■\begin{align*} & \textrm{Greatest possible } \textrm{P}\left(A' \cap B' \cap C'\right) \\ &\quad= 0.33 \; \blacksquare \\ & \textrm{Least possible } \textrm{P}\left(A' \cap B' \cap C'\right) \\ &\quad= 0.165 \; \blacksquare \end{align*}