Sets, Venn diagrams and sample spaces

(a) $\{6, 12\}$

1. $A = \{3, 6, 9, 12\}$ and $B = \{2, 4, 6, 8, 10, 12\}$.
2. $A \cap B$ (intersection) is the numbers in both: 6 and 12.

• B1 Correct answer: 6 and 12 only.

(b) 8

1. $A \cup B$ is everything in $A$ or $B$ or both: $\{2, 3, 4, 6, 8, 9, 10, 12\}$.

• B1 Correct answer: 8.

(c) $\frac{2}{3}$

1. $A'$ is the complement: numbers not in $A$. There are $12 - 4 = 8$ of them.
2. $P(A') = \frac{8}{12} = \frac{2}{3}$.

• M1 Counting 8 numbers not in $A$.
• A1 $\frac{8}{12}$ or equivalent.

(a) $\frac{1}{6}$

1. A sample space grid has $6 \times 6 = 36$ equally likely outcomes.
2. Totals of 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1): 6 outcomes. $P = \frac{6}{36}$.

• M1 36 outcomes, or listing the 6 pairs that total 7.
• A1 $\frac{6}{36}$ or equivalent.

(b) $\frac{3}{36}$

1. Totals of 11: (5, 6), (6, 5). Total of 12: (6, 6). That is 3 outcomes.

• B1 $\frac{3}{36}$ or equivalent.

(a) $6$

1. $$25+19-8$$
2. $$42-36$$
3. Use inclusion and exclusion: number studying at least one language = 25 + 19 $-$ 8 = 36.
4. Subtract from the whole group: neither = 42 $-$ 36 = 6.

• P1 Establishing $25+19-8$ or an equivalent valid method.
• P1 Establishing $42-36$ or an equivalent valid method.
• A1 Correct answer: $6$

(a) 1/2

1. $$1/3\times 2/4+2/3\times 2/4$$
2. There are 3 $\times$ 4 = 12 equally likely pairs.
3. An odd sum needs one odd and one even number. The first set has 1 odd and 2 even numbers; the second has 2 of each.
4. Favourable pairs = 1 $\times$ 2 + 2 $\times$ 2 = 6. Therefore the probability is 6/12 = 1/2.

• P1 Establishing $1/3\times 2/4+2/3\times 2/4$ or an equivalent valid method.
• A1 Correct answer: 1/2

(a) 1/3

1. $$2/4\times 1/3+1/4\times 2/3$$
2. Treat the two cards labelled 1 as separate cards. There are 4 $\times$ 3 = 12 equally likely ordered draws.
3. A total of 4 requires a 1 and a 3. There are 2 $\times$ 1 = 2 orders starting with a 1 and 1 $\times$ 2 = 2 orders starting with the 3.
4. There is only one card labelled 2, so (2,2) is impossible. Probability = 4/12 = 1/3.

• P1 Establishing $2/4\times 1/3+1/4\times 2/3$ or an equivalent valid method.
• A1 Correct answer: 1/3

(a) $16$

1. Use inclusion-exclusion to count the union once.
   $$24+20-10$$
2. Subtract the union from the group total.
   $$50-34$$
3. Therefore $16$.

• M1 Use inclusion-exclusion to count the union once.
• M1 Subtract the union from the group total.
• A1 Correct answer: $16$

(b) $12/25$

1. Remove the overlap from each drink total.
   $$\frac{24-10+20-10}{50}$$
2. Therefore $12/25$.

• M1 Remove the overlap from each drink total.
• A1 Correct answer: $12/25$

(a) $3, 6, 15$

1. Therefore $3, 6, 15$.

• B1 Correct answer: $3, 6, 15$

(b) $9$

1. Add the sizes of A and B, then subtract the overlap once.
   $$5+7-3$$
2. Therefore $9$.

• M1 Add the sizes of A and B, then subtract the overlap once.
• A1 Correct answer: $9$

(c) $2/5$

1. Count the numbers outside A ∪ B.
   $$\frac{15-9}{15}$$
2. Therefore $2/5$.

• M1 Count the numbers outside A ∪ B.
• A1 Correct answer: $2/5$

(a) $\frac{14}{25}$

1. All four regions add up to 50.
   $$2x + x + (3x - 4) + 6 = 50$$
2. $$6x + 2 = 50 \;\Rightarrow\; x = 8$$
3. Hockey: both $+$ hockey only $= 8 + 20 = 28$.
4. $P(H) = \frac{28}{50} = \frac{14}{25}$.

• P1 An equation adding the four regions to 50.
• P1 Solving to $x = 8$.
• P1 Hockey total 28 (both plus hockey only).
• A1 $\frac{28}{50}$ or equivalent.

(b) $\frac{6}{50}$

1. $(F \cup H)'$ is outside both circles: the 6 who play neither.

• B1 $\frac{6}{50}$ or equivalent.