2018 H2 Maths Paper 2 Q7 — Worked Solution

Part (i)

2 marks

Since $A$ and $B$ are independent, $P(A \cap B) = ab$ .

P(A'\cap B')

= 1 - P(A \cup B)

= 1 - \left\lbrack P(A) + P(B) - P(A \cap B) \right\rbrack

= 1 - a - b + ab

= (1 - a)(1 - b).

Since $P(A') = 1 - a$ and $P(B') = 1 - b$ , this says

P(A'\cap B') = P(A')\, P(B'),

so $A'$ and $B'$ are independent events. (shown)

Part (ii)

2 marks

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

P(A'\cap C')

= 1 - P(A \cup C)

= 1 - \left\lbrack P(A) + P(C) - P(A \cap C) \right\rbrack

= 1 - a - c.

If $A'$ and $C'$ are also mutually exclusive, then $P(A'\cap C') = 0$ , i.e. $a + c = 1$ . Together with $A \cap C = \varnothing$ , this means every outcome in the sample space lies in exactly one of $A$ or $C$ — the two events partition the universal set completely, with no overlap and no region left outside both.

Part (iii)

4 marks

Approach

$A$ and $C$ are mutually exclusive, so $A$ , $C$ and $N: = (A \cup C)'$ partition the sample space into three disjoint pieces. Splitting $B$ across this partition, $P(B) = P(A \cap B) + P(C \cap B) + P(N \cap B)$ . This turns the two given pieces of data — $P(B \cap C)$ and $P(A'\cap B'\cap C') = P(N \cap B')$ — together with $A,B$ independence, into two equations relating $P(A \cap B)$ to $c = P(C)$ , the one remaining unknown. The extreme values of $P(A \cap B)$ then come from the extreme values of $c$ allowed by the constraints of the problem (each probability lying in $\lbrack 0,1\rbrack$ ).

We are given $a = P(A) = \frac{2}{5}$ , $P(B \cap C) = \frac{1}{5}$ and $P(A'\cap B'\cap C') = \frac{1}{10}$ . Write $c = P(C)$ and $b = P(B)$ , both unknown for now, and let $N = (A \cup C)'$ , so $P(N) = 1 - a - c = \frac{3}{5} - c$ .

Since $A \cap C = \varnothing$ , the events $A \cap B$ , $C \cap B$ and $N \cap B$ partition $B$ :

b = P(A \cap B) + P(C \cap B) + P(N \cap B).

Using $A,B$ independence, $P(A \cap B) = ab = \frac{2}{5}b$ . Writing $z = P(N \cap B)$ ,

b = \frac{2}{5}b + \frac{1}{5} + z.

Also, $P(A'\cap B'\cap C') = P(N \cap B') = P(N) - P(N \cap B) = \left( \frac{3}{5} - c \right) - z$ , and this equals $\frac{1}{10}$ :

\left( \frac{3}{5} - c \right) - z = \frac{1}{10}

\Rightarrow z = \frac{1}{2} - c.

Substituting into the equation for $b$ :

b = \frac{2}{5}b + \frac{1}{5} + \left( \frac{1}{2} - c \right)

\Rightarrow \frac{3}{5}b = \frac{7}{10} - c

\Rightarrow b = \frac{7}{6} - \frac{5}{3}c.

Therefore

P(A \cap B) = \frac{2}{5}b = \frac{2}{5}\left( \frac{7}{6} - \frac{5}{3}c \right) = \frac{7}{15} - \frac{2}{3}c,

a decreasing function of $c$ , so the maximum of $P(A \cap B)$ occurs at the smallest possible $c$ , and the minimum at the largest possible $c$ .

We now bound $c$ . Since $B \cap C \subseteq C$ , $c \geq P(B \cap C) = \frac{1}{5}$ . Since $z = P(N \cap B)$ is a probability, $z \geq 0$ , i.e. $c \leq \frac{1}{2}$ . (Checking the remaining probabilities — $b$ , $P(A \cap B)$ , $z \leq P(N)$ — against these two bounds shows none of them cuts the range further, and $a + c \leq \frac{2}{5} + \frac{1}{2} < 1$ throughout, so $A'$ and $C'$ staying not mutually exclusive is automatic.)

\frac{1}{5} \leq c \leq \frac{1}{2}.

At $c = \frac{1}{5}$ : $P(A \cap B) = \frac{7}{15} - \frac{2}{3} \cdot \frac{1}{5} = \frac{7}{15} - \frac{2}{15} = \frac{5}{15} = \frac{1}{3}$ .

At $c = \frac{1}{2}$ : $P(A \cap B) = \frac{7}{15} - \frac{2}{3} \cdot \frac{1}{2} = \frac{7}{15} - \frac{1}{3} = \frac{7}{15} - \frac{5}{15} = \frac{2}{15}$ .

To check your answer: at

c = \frac{1}{5}

,

b = \frac{7}{6} - \frac{5}{3} \cdot \frac{1}{5} = \frac{5}{6}

, and

P(A \cap B) = \frac{1}{3} \leq \min(a,b) = \frac{2}{5}

, a valid probability; at

c = \frac{1}{2}

,

b = \frac{1}{3}

and

P(A \cap B) = \frac{2}{15} \leq \frac{2}{5}

also. ✓

$\therefore$ the maximum possible value of $P(A \cap B)$ is $\frac{1}{3}$ , and the minimum possible value is $\frac{2}{15}$ .

2018 H2 Maths Paper 2, Question 7

Part (i)

Part (ii)

Part (iii)

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