2010 H2 Mathematics Paper 2 Question 7

Probability

Answers

(i)

0.32{0.32}

(ii)

0.92{0.92}

(iii)

1635{\frac{16}{35}}

(iv)

0.15{0.15}

(v)

P(A′∩B∩C)≤0.15{\textrm{P}\left(A' \cap B \cap C\right) \leq 0.15}

Full solutions

(i)

P(A∣B′)=0.8P(A∩B′)P(B′)=0.8P(A∩B′)=0.8×(1−P(B))P(A∩B′)=0.8×(1−0.6)P(A∩B′)=0.32  ■\begin{align*} \textrm{P}\left(A \mid B'\right) &= 0.8 \\ \frac{\textrm{P}\left(A \cap B'\right)}{\textrm{P}\left(B'\right)} &= 0.8 \\ \textrm{P}\left(A \cap B'\right) &= 0.8 \times \Big(1 - \textrm{P}\left(B\right)\Big) \\ \textrm{P}\left(A \cap B'\right) &= 0.8 \times \left(1 - 0.6\right) \\ \textrm{P}\left(A \cap B'\right) &= 0.32 \; \blacksquare \end{align*}

(ii)

P(A∪B)=P(A∩B′)+P(B)=0.32+0.6=0.92  ■\begin{align*} \textrm{P}\left(A \cup B\right) &= \textrm{P}\left(A\cap B'\right) + \textrm{P}\left(B\right) \\ &= 0.32 + 0.6 \\ &= 0.92 \; \blacksquare \end{align*}

(iii)

P(B′∣A)=P(B′∩A)P(A)=0.320.7=1635  ■\begin{align*} \textrm{P}\left(B' \mid A\right) &= \frac{\textrm{P}\left(B' \cap A\right)}{\textrm{P}\left(A\right)} \\ & = \frac{0.32}{0.7} \\ & = \frac{16}{35} \; \blacksquare \end{align*}

(iv)

Since A{A} and C{C} are independent,
P(A∩C)=P(A)⋅P(C)\textrm{P}\left(A \cap C\right) = \textrm{P}\left(A\right) \cdot \textrm{P}\left(C\right)
P(A′∩C)=P(C)−P(A∩C)=P(C)−P(A)⋅P(C)=0.5−0.7(0.5)=0.15  ■\begin{align*} \textrm{P}\left(A'\cap C\right) &= \textrm{P}\left(C\right) - \textrm{P}\left(A \cap C\right) \\ &= \textrm{P}\left(C\right) - \textrm{P}\left(A\right) \cdot \textrm{P}\left(C\right) \\ & = 0.5 - 0.7(0.5) \\ & = 0.15 \; \blacksquare \end{align*}

(v)

Since (A′∩B∩C)⊆(A′∩C),{(A' \cap B \cap C) \subseteq (A' \cap C),}
P(A′∩B∩C)≤P(A∩C)=0.15  ■\begin{align*} \textrm{P}\left(A' \cap B \cap C\right) &\leq \textrm{P}\left(A \cap C\right) \\ &= 0.15 \; \blacksquare \end{align*}