2019 H2 Maths Paper 2 Q8 — Worked Solution

We first find the row and column totals of the table (each is the number of items of that character, or of that colour):

	Orange	Yellow	Green	White	Total
Horse	1	1	3	4	9
Rider	1	1	7	5	14
Dog	3	7	1	6	17
Bird	4	5	6	1	16
Total	9	14	17	16	56

Part (i)(a)

1 mark

P\left( \text{Horse or Rider} \right) = \frac{9 + 14}{56}

= \frac{23}{56}

Part (i)(b)

1 mark

Approach

“Dog or Bird but not White” removes the White items from the Dog and Bird rows, since a Dog or Bird item that is White is excluded.

P\left( \text{Dog or Bird, not White} \right) = \frac{(17 - 6) + (16 - 1)}{56}

= \frac{11 + 15}{56}

= \frac{26}{56}

= \frac{13}{28}

Part (ii)(a)

1 mark

Approach

choosing two items at random (without replacement) from the bag, we want both to lie in the “Horse, not Orange” cells, i.e. among the $9 - 1 = 8$ items that are a Horse and not Orange.

P\left( \text{both Horses, neither Orange} \right) = \frac{\binom{8}{2}}{\binom{56}{2}}

= \frac{28}{1540}

= \frac{1}{55}

Part (ii)(b)

3 marks

Approach

the two chosen items must together contain exactly one Dog and exactly one Yellow item. Since a single item can be both a Dog and Yellow at once (the Dog–Yellow cell, 7 items), we split by whether this double-counting cell is used: either one item is Dog-and-Yellow and the other is neither a Dog nor Yellow (this alone already gives exactly one Dog and one Yellow), or one item is a Dog that is not Yellow and the other is Yellow but not a Dog. Any other combination gives too many Dogs or too many Yellow items.

From the table: Dog-and-Yellow items number $7$ ; Dog-but-not-Yellow items number $17 - 7 = 10$ ; Yellow-but-not-Dog items number $14 - 7 = 7$ ; and the remaining items (neither Dog nor Yellow) number $56 - 7 - 10 - 7 = 32$ .

P\left( \text{exactly 1 Dog and exactly 1 Yellow} \right) = \frac{\binom{10}{1}\binom{7}{1} + \binom{7}{1}\binom{32}{1}}{\binom{56}{2}}

= \frac{70 + 224}{1540}

= \frac{294}{1540}

= \frac{21}{110}

To check your answer:

70 + 224 = 294

favourable outcomes out of

\binom{56}{2} = 1540

total, and

294/1540

simplifies (dividing by

14

) to

21/110

. ✓

Part (iii)

3 marks

Approach

if Gerri’s two favourite combinations are the cells with counts $m$ and $n$ , choosing one item of each favourite (in either order) from the bag of 56 items gives probability $\frac{mn}{\binom{56}{2}}$ . Setting this equal to $\frac{1}{77}$ pins down the product $mn$ , which we then match against the counts actually in the table.

\frac{mn}{\binom{56}{2}} = \frac{1}{77}

mn = \frac{1540}{77}

mn = 20

The 16 cell counts in the table are $1,\, 3,\, 4,\, 5,\, 6,\, 7$ (with repeats). The only pair of these values with product $20$ is $4 \times 5$ .

The cells with count $4$ are Horse–White and Bird–Orange; the cells with count $5$ are Rider–White and Bird–Yellow. Gerri’s two favourites are one cell from each list:

\therefore\ \text{Gerri’s two favourites are one of:}

Horse (White) and Rider (White)
Horse (White) and Bird (Yellow)
Bird (Orange) and Rider (White)
Bird (Orange) and Bird (Yellow)

2019 H2 Maths Paper 2, Question 8

Part (i)(a)

Part (i)(b)

Part (ii)(a)

Part (ii)(b)

Part (iii)

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