2015 H2 Maths Paper 2 Q9 — Worked Solution

Part (i)

1 mark

Since $A$ and $B$ are independent,

P(B \mid A) = P(B) = 0.4

Part (ii)

3 marks

Approach

we want $P(A'\cap B'\cap C')$ , the probability that none of the three events occur. It is easiest to find this as $1 - P(A \cup B \cup C)$ , and $P(A \cup B \cup C)$ is found by inclusion-exclusion. Since $A,B$ are independent, $A,C$ are independent, and now $B,C$ are independent too, every pairwise intersection is just a product of probabilities.

P(A \cap B) = P(A)P(B) = 0.45 \times 0.4 = 0.18

P(A \cap C) = P(A)P(C) = 0.45 \times 0.3 = 0.135

P(B \cap C) = P(B)P(C) = 0.4 \times 0.3 = 0.12

By inclusion-exclusion,

P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)

= 0.45 + 0.4 + 0.3 - 0.18 - 0.135 - 0.12 + 0.1

= 0.815

Note: even though every pair of events is independent here, $P(A \cap B \cap C) = 0.1 \neq P(A)P(B)P(C) = 0.054$ , so the three events are not mutually independent — the given value of $0.1$ has to be used as it stands, not replaced by a product.

P(A'\cap B'\cap C') = 1 - P(A \cup B \cup C)

= 1 - 0.815

= 0.185

$\therefore$ $P(A'\cap B'\cap C') = 0.185$ .

Part (iii)

4 marks

Approach

with $B$ and $C$ no longer independent, $P(B \cap C)$ is not pinned down by $P(B)$ and $P(C)$ alone; it can take a range of values, and $P(A)$ , $P(A \cap B \cap C) = 0.1$ , and the independence of $A$ with $B$ and with $C$ stay fixed throughout. We split the sample space into the seven regions of the Venn diagram, express each region’s probability in terms of $x = P(B \cap C)$ , and require every region to be non-negative — this pins down the range of $x$ , and hence of $P(A'\cap B'\cap C')$ .

Since $A,B$ are independent and $A,C$ are independent, $P(A \cap B) = 0.18$ and $P(A \cap C) = 0.135$ as before, and $P(A \cap B \cap C) = 0.1$ still. So

P(A \cap B \cap C') = P(A \cap B) - P(A \cap B \cap C) = 0.18 - 0.1 = 0.08

P(A \cap B'\cap C) = P(A \cap C) - P(A \cap B \cap C) = 0.135 - 0.1 = 0.035

P\left( A\ \text{only} \right) = P(A) - P(A \cap B) - P(A \cap C) + P(A \cap B \cap C) = 0.45 - 0.18 - 0.135 + 0.1 = 0.235

Let $x = P(B \cap C)$ . Then

P(A'\cap B \cap C) = x - 0.1

P\left( B\ \text{only} \right) = P(B) - P(A \cap B) - x + P(A \cap B \cap C) = 0.4 - 0.18 - x + 0.1 = 0.32 - x

P\left( C\ \text{only} \right) = P(C) - P(A \cap C) - x + P(A \cap B \cap C) = 0.3 - 0.135 - x + 0.1 = 0.265 - x

Each of these seven regional probabilities must be non-negative. $P(A'\cap B \cap C) = x - 0.1 \geq 0$ gives $x \geq 0.1$ ; $P\left( B\ \text{only} \right) = 0.32 - x \geq 0$ gives $x \leq 0.32$ ; $P\left( C\ \text{only} \right) = 0.265 - x \geq 0$ gives $x \leq 0.265$ . Together,

0.1 \leq x \leq 0.265

Summing all seven regions,

P(A \cup B \cup C) = 0.235 + (0.32 - x) + (0.265 - x) + 0.08 + 0.035 + (x - 0.1) + 0.1

= 0.935 - x

so

P(A'\cap B'\cap C') = 1 - (0.935 - x)

= 0.065 + x

This is increasing in $x$ , so it is greatest when $x$ is greatest and least when $x$ is least.

At $x = 0.265$ : $P(A'\cap B'\cap C') = 0.065 + 0.265 = 0.33$ . At $x = 0.1$ : $P(A'\cap B'\cap C') = 0.065 + 0.1 = 0.165$ .

To check your answer: at

x = 0.1

,

P(A'\cap B \cap C) = 0

, i.e.

B \cap C

occurs only together with

A

— a valid boundary case; at

x = 0.265

,

P\left( C\ \text{only} \right) = 0

, another valid boundary case. Both extremes correspond to one region vanishing, as expected at the edge of a feasible range. ✓

$\therefore$ the greatest possible value of $P(A'\cap B'\cap C')$ is $0.33$ , and the least possible value is $0.165$ .

2015 H2 Maths Paper 2, Question 9

Part (i)

Part (ii)

Part (iii)

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