2013 H2 Mathematics Paper 2 Question 8

Probability

Answers

(i)

0.24{0.24}

(ii)

0.06{0.06}

(iii)

0.26{0.26}

Full solutions

(i)

P(B∣A′)=0.8P(B∩A′)P(A′)=0.8P(B∩A′)=0.8×(1−P(A))P(B∩A′)=0.8×(1−0.7)P(B∩A′)=0.24  ■\begin{align*} \textrm{P}\left(B \mid A'\right) &= 0.8 \\ \frac{\textrm{P}\left(B \cap A'\right)}{\textrm{P}\left(A'\right)} &= 0.8 \\ \textrm{P}\left(B \cap A'\right) &= 0.8 \times \Big(1 - \textrm{P}\left(A\right)\Big) \\ \textrm{P}\left(B \cap A'\right) &= 0.8 \times \left(1 - 0.7\right) \\ \textrm{P}\left(B \cap A'\right) &= 0.24 \; \blacksquare \end{align*}

(ii)

P(A∪B)=P(B∩A′)+P(A)=0.24+0.7=0.94\begin{align*} \textrm{P}\left(A \cup B\right) &= \textrm{P}\left(B\cap A'\right) + \textrm{P}\left(A\right) \\ &= 0.24 + 0.7 \\ &= 0.94 \end{align*}
P(A′∩B′)=1−P(A∪B)=1−0.94=0.06  ■\begin{align*} \textrm{P}\left(A' \cap B'\right) &= 1 - \textrm{P}\left(A \cup B\right) \\ &= 1 - 0.94 \\ &= 0.06 \; \blacksquare \end{align*}

(iii)

Let x{x} denote P(A∩B){\textrm{P}\left(A \cap B\right)}
P(A∣B′)=0.88P(A∩B′)P(B′)=0.880.7−x1−(0.24+x)=0.88\begin{align*} \textrm{P}\left(A \mid B'\right) &= 0.88 \\ \frac{\textrm{P}\left(A \cap B'\right)}{\textrm{P}\left(B'\right)} &= 0.88 \\ \frac{0.7-x}{1-(0.24+x)} &= 0.88 \\ \end{align*}
Hence P(A∩B)=x=0.26  ■{\textrm{P}\left(A \cap B\right) = x = 0.26 \; \blacksquare}