<p>The events $A$, $B$ and $C$ are such that $P(A)=a$, $\mathrm{P}\left( B \right)=b$ and $\mathrm{P}\left( C \right)=c$. $A$ and $B$ are independent events. $A$ and $C$ are mutually exclusive events.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Find an expression for $\mathrm{P}\left( A'\cap B' \right)$ and hence prove that $A'$ and $B'$ are independent events.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Find an expression for $\mathrm{P}\left( A'\cap C' \right)$. Draw a Venn diagram to illustrate to case when $A'$ and $C'$ are also mutually exclusive events. (You should not show event $B$on your diagram.)</div><p style="text-align: right;">[2]</p></li></ol>
<p>You are now given that $A'$ and $C'$ are not mutually exclusive, $P(A)=\frac{2}{5}$, $\mathrm{P}\left( B\cap C \right)=\frac{1}{5}$ and $\mathrm{P}\left( A'\cap B'\cap C' \right)=\frac{1}{10}$.</p>
<ol class="i-list" start="3"><li><div class="list-item-paragraph">Find exactly the maximum and minimum possible values of $\mathrm{P}\left( A\cap B \right)$.</div><p style="text-align: right;">[4]</p></li></ol>

The events $A$, $B$ and $C$ are such that $P(A)=a$, $\mathrm{P}\left( B \right)=b$ and $\mathrm{P}\left( C \right)=c$. $A$ and $B$ are independent events. $A$ and $C$ are mutually exclusive events.

Find an expression for $\mathrm{P}\left( A'\cap B' \right)$ and hence prove that $A'$ and $B'$ are independent events.

Find an expression for $\mathrm{P}\left( A'\cap C' \right)$. Draw a Venn diagram to illustrate to case when $A'$ and $C'$ are also mutually exclusive events. (You should not show event $B$on your diagram.)

You are now given that $A'$ and $C'$ are not mutually exclusive, $P(A)=\frac{2}{5}$, $\mathrm{P}\left( B\cap C \right)=\frac{1}{5}$ and $\mathrm{P}\left( A'\cap B'\cap C' \right)=\frac{1}{10}$.

Find exactly the maximum and minimum possible values of $\mathrm{P}\left( A\cap B \right)$.

Probability - General

Probability

International Baccalaureate Higher Level Mathematics

IB HL Math

N2018 P2 Q7

$P(A) = a$, $P(B) = b$, $P(C) = c$

$P(A \cap B) = P(A) \cdot P(B)$, $P(A \cap C) = 0$

Need to show: $P(A' \cap B') = P(A') \cdot P(B')$

LHS $= P(A' \cap B')$

$= 1 - P(A \cup B)$

$= 1 - [P(A) + P(B) - P(A \cap B)]$

$= 1 - [a + b - ab]$

$= 1 - a - b + ab$

$= (1 - a) - b(1 - a)$

$= (1 - a)(1 - b)$

$= P(A') \cdot P(B')$ (shown)

$\therefore A'$ and $B'$ are independent events.

$P(A' \cap C') = 1 - a - c$

$P(A \cap C) = 0$, since $A$ and $C$ are mutually exclusive.

$\Rightarrow P(A' \cap C') = 0 \Rightarrow 1 - a - c = 0$

$a + c = 1$

From Venn diagram: $P(A) = \dfrac{2}{5} = 0.4$, $P(B \cap C) = \dfrac{1}{5} = 0.2$, $P(A \cup B \cup C) = 0.9$

Let $P(B \cap A' \cap C') = y$ and $P(B' \cap C) = x$

$0.4 + y + 0.2 + x = 0.9$

$y + x = 0.3$

Max of $y = 0.3$, min of $y = 0$

$P(A \cap B) = P(A) \cdot P(B) = 0.4[P(A \cap B) + y + 0.2]$

$P(A \cap B) = 0.4 \cdot P(A \cap B) + 0.4y + 0.08$

$0.6 \cdot P(A \cap B) = 0.4y + 0.08$

$P(A \cap B) = \dfrac{2}{3}y + \dfrac{2}{15}$

Max of $P(A \cap B) = \dfrac{2}{3}(0.3) + \dfrac{2}{15} = \dfrac{1}{3}$

Min of $P(A \cap B) = \dfrac{2}{3}(0) + \dfrac{2}{15} = \dfrac{2}{15}$

$\therefore \dfrac{2}{15} \leq P(A \cap B) \leq \dfrac{1}{3}$

N2018 P2 Q7

N2018 P2 Q7

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