2019 H2 Maths Paper 2 Q7 — Worked Solution

Part (i)

2 marks

Two assumptions needed to model $F$ by a binomial distribution:

Whether any one mug is faulty is independent of whether any other mug is faulty.
Every mug has the same, constant probability $0.08$ of being faulty (e.g. the production process does not change over the period the sample is collected).

Part (ii)

2 marks

Let $F$ be the number of faulty mugs found in a randomly chosen sample of 50. Then

F \sim B(50,\ 0.08)

Note: a binomial tail is evaluated by converting to cumulative form first.

P(F \geq 7) = 1 - P(F \leq 6)

By GC, $P(F \leq 6) = 0.898$ (unrounded $0.898128$ ), so

P(F \geq 7) = 1 - 0.898128

= 0.102\ \text{(3 s.f.)}

Part (iii)

2 marks

Approach

each of the 5 days of the week independently either does or does not have “at least 7 faulty mugs found”, with probability $P(F \geq 7)$ found in (ii). This is itself a new binomial variable, counting a binomial number of “successful” days.

Let $W$ be the number of days, out of a working week of 5 days, on which at least 7 faulty mugs are found. Since the number of faulty mugs on different days is independent, and each day has probability $P(F \geq 7) = 0.101872$ (unrounded value from (ii)) of being such a day,

W \sim B(5,\ 0.101872)

P(W \leq 2) = 0.991\ \text{(3 s.f.)}

by GC (unrounded $0.990978$ ).

Part (iv)

1 mark

Let $S$ be the number of faulty saucers in a random sample of 10, so $S \sim B(10,\ p)$ .

P(S = 2) = \binom{10}{2}p^{2}(1 - p)^{8}

= 45p^{2}(1 - p)^{8}

Part (v)

4 marks

Approach

a set has 2 mugs and 2 saucers, so the number of faulty items in a set is the sum of a $B(2,\, 0.08)$ variable (faulty mugs) and an independent $B(2,\, p)$ variable (faulty saucers). “At most 1 faulty item” is the event that this sum is $0$ or $1$ ; we split into the cases 0 faulty mugs & 0 or 1 faulty saucer, or 1 faulty mug & 0 faulty saucers.

Let $M$ be the number of faulty mugs and $S'$ the number of faulty saucers in a set, so $M \sim B(2,\, 0.08)$ and $S'\sim B(2,\, p)$ independently.

P(M = 0) = {0.92}^{2} = 0.8464,\quad\quad P(M = 1) = 2(0.08)(0.92) = 0.1472

P\left( \text{at most 1 faulty item} \right) = P(M = 0)\left\lbrack P(S'= 0) + P(S'= 1) \right\rbrack + P(M = 1)\, P(S'= 0)

Since $P(S'= 0) + P(S'= 1) = (1 - p)^{2} + 2p(1 - p) = (1 - p)(1 + p) = 1 - p^{2}$ ,

0.8464\left( 1 - p^{2} \right) + 0.1472(1 - p)^{2} = 0.97

Solving by GC, the equation has two roots, $p = 0.0689$ or $p = - 0.490$ ; since $p$ is a probability, $0 < p < 1$ ,

p = - 0.490\ \left( \text{rej. as }p > 0 \right)

\therefore\ p = 0.0689\ \text{(3 s.f.)}

To check your answer: with

p = 0.0689

,

0.8464\left( 1 - p^{2} \right) + 0.1472(1 - p)^{2} = 0.970

, matching the given

0.97

. ✓

2019 H2 Maths Paper 2, Question 7

Part (i)

Part (ii)

Part (iii)

Part (iv)

Part (v)

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