Algebraic probability and linked events

(a) 0.3

1. $$2x+3x+x+0.1+0.3=1$$
2. $$6x=0.6$$
3. The probabilities add to 1: 2x + 3x + x + 0.1 + 0.3 = 1.
4. Therefore 6x = 0.6, so x = 0.1.
5. P(B) = 3x = 0.3.

• P1 Establishing $2x+3x+x+0.1+0.3=1$ or an equivalent valid method.
• P1 Establishing $6x=0.6$ or an equivalent valid method.
• A1 Correct answer: 0.3

(a) Yes: $P(A) \times P(B) = 0.2 = P(A \cap B)$.

1. Independent means $P(A \cap B) = P(A) \times P(B)$.
2. $0.4 \times 0.5 = 0.2$, which matches.

• M1 $0.4 \times 0.5 = 0.2$ compared with $P(A \cap B)$ (or $P(A \mid B) = \frac{0.2}{0.5} = 0.4 = P(A)$).
• C1 "Yes" with the comparison stated.

(b) 0.7

1. $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.5 - 0.2$.

• B1 Correct answer: 0.7.

(a) $21$

1. $$(30-x)/(54-2x)=5/8$$
2. $$8(30-x)=5(54-2x)$$
3. $$60-(30+24-15)$$
4. Let x attend both clubs. Music only = 30 $-$ x; sports only = 24 $-$ x.
5. Conditioning on exactly one club gives (30 $-$ x)/(54 $-$ 2x) = 5/8.
6. Cross-multiply: 240 $-$ 8x = 270 $-$ 10x, so x = 15.
7. By inclusion and exclusion, 30 + 24 $-$ 15 = 39 attend at least one club. Therefore 60 $-$ 39 = 21 attend neither.

• P1 Establishing $(30-x)/(54-2x)=5/8$ or an equivalent valid method.
• P1 Establishing $8(30-x)=5(54-2x)$ or an equivalent valid method.
• P1 Establishing $60-(30+24-15)$ or an equivalent valid method.
• A1 Correct answer: $21$

(a) $7$

1. Model the dependent two-draw probability.
   $$3/(n+3)\times 2/(n+2)=1/15$$
2. Clear denominators and factorise the quadratic.
   $$(n-7)(n+12)=0$$
3. Therefore $7$.

• P1 Model the dependent two-draw probability.
• P1 Clear denominators and factorise the quadratic.
• A1 Correct answer: $7$

(b) $7/15$

1. Add the green-orange and orange-green routes using n=7.
   $$7/10\times 3/9+3/10\times 7/9$$
2. Therefore $7/15$.

• P1 Add the green-orange and orange-green routes using n=7.
• A1 Correct answer: $7/15$

(a) $\frac{6}{n} \times \frac{5}{n - 1} = \frac{1}{3}$ gives $n^2 - n = 90$.

1. $$\frac{6}{n} \times \frac{5}{n - 1} = \frac{1}{3}$$
2. $$\frac{30}{n(n - 1)} = \frac{1}{3} \;\Rightarrow\; 90 = n^2 - n$$
3. So $n^2 - n - 90 = 0$.

• P1 $\frac{6}{n} \times \frac{5}{n - 1}$.
• P1 Setting it equal to $\frac{1}{3}$ and clearing the fractions.
• A1 Reaching $n^2 - n - 90 = 0$ with every step shown.

(b) 10

1. Factorise.
   $$(n - 10)(n + 9) = 0$$
2. $n$ is a number of sweets, so $n = 10$.

• M1 $(n - 10)(n + 9)$ or the quadratic formula.
• A1 Correct answer: 10.

(a) 4

1. Different colours: red then blue, or blue then red.
   $$2 \times \frac{x}{x + 5} \times \frac{5}{x + 4} = \frac{5}{9}$$
2. $$\frac{10x}{(x + 5)(x + 4)} = \frac{5}{9} \;\Rightarrow\; 90x = 5(x^2 + 9x + 20)$$
3. $$x^2 - 9x + 20 = 0 \;\Rightarrow\; (x - 4)(x - 5) = 0$$
4. $x = 4$ or $x = 5$. There are more blue than red, so $x = 4$.

• P1 One product $\frac{x}{x + 5} \times \frac{5}{x + 4}$.
• P1 Doubling for both orders and setting equal to $\frac{5}{9}$.
• P1 Reaching $x^2 - 9x + 20 = 0$.
• P1 Solving: $x = 4$ or $x = 5$.
• A1 $x = 4$, using the fact that there are fewer red counters than blue.

(a) $0.3$

1. $$(1-p)(1-2p)=0.28$$
2. $$2p^{2}-3p+0.72=0$$
3. The miss probabilities are 1 $-$ p and 1 $-$ 2p. Independence gives (1 $-$ p)(1 $-$ 2p) = 0.28.
4. Expand and rearrange: $2p^{2}$ $-$ 3p + 0.72 = 0, so $50p^{2}$ $-$ 75p + 18 = 0.
5. Factorise: (10p $-$ 3)(5p $-$ 6) = 0, giving p = 0.3 or p = 1.2.
6. Probabilities must lie from 0 to 1, and 2p $\le$ 1. Therefore p = 0.3. Check: 0.7 $\times$ 0.4 = 0.28.

• P1 Establishing $(1-p)(1-2p)=0.28$ or an equivalent valid method.
• P1 Establishing $2p^{2}-3p+0.72=0$ or an equivalent valid method.
• A1 Correct answer: $0.3$

(a) $6$

1. $$2n\times 4/((n+4)(n+3))=8/15$$
2. $$n^{2}-8n+12=0$$
3. Different colours can occur in either order: P(different) = n/(n + 4) $\times$ 4/(n + 3) + 4/(n + 4) $\times$ n/(n + 3).
4. Set 8n/[(n + 4)(n + 3)] = 8/15. Simplifying gives 15n = (n + 4)(n + 3).
5. Rearrange: $n^{2}$ $-$ 8n + 12 = 0, so (n $-$ 2)(n $-$ 6) = 0.
6. Since n > 4, n = 6. Check: 2 $\times$ 6/10 $\times$ 4/9 = 8/15.

• P1 Establishing $2n\times 4/((n+4)(n+3))=8/15$ or an equivalent valid method.
• P1 Establishing $n^{2}-8n+12=0$ or an equivalent valid method.
• A1 Correct answer: $6$