Conditional probability in tables, trees and Venn diagrams

(a) 4/11

1. $$12+21$$
2. $$12/33$$
3. Condition on the evening performance: there are 12 + 21 = 33 eligible tickets.
4. Of these, 12 are child tickets, so the probability is 12/33 = 4/11.

• P1 Establishing $12+21$ or an equivalent valid method.
• P1 Establishing $12/33$ or an equivalent valid method.
• A1 Correct answer: 4/11

(a) 14/25

1. $$46-18$$
2. $$80-30$$
3. $$28/50$$
4. Cyclists who do not swim: 46 $-$ 18 = 28.
5. Non-cyclists total 80 $-$ 46 = 34; non-cyclists who do not swim total 34 $-$ 12 = 22.
6. Condition on not swimming: probability = 28/(28 + 22) = 14/25.

• P1 Establishing $46-18$ or an equivalent valid method.
• P1 Establishing $80-30$ or an equivalent valid method.
• P1 Establishing $28/50$ or an equivalent valid method.
• A1 Correct answer: 14/25

(a) $\frac{8}{23}$

1. Dog owners: $8 + 15 = 23$.
2. Of these, 8 also own a cat: $P(A \mid B) = \frac{8}{23}$.

• M1 Denominator 23, the number of dog owners.
• A1 Correct answer: $\frac{8}{23}$.

(b) $\frac{15}{35}$

1. $A \cup B$ has $12 + 8 + 15 = 35$ people.
2. Of these, those not owning a cat are the 15 dog-only owners.

• M1 Using the denominator 35.
• A1 $\frac{15}{35}$ or equivalent.

(a) 9/25

1. $$3/5\times 1/4$$
2. $$2/5\times 2/3$$
3. $$(3/20)/(3/20+4/15)$$
4. Multiply along each path: P(A and red) = 3/5 $\times$ 1/4 = 3/20; P(B and red) = 2/5 $\times$ 2/3 = 4/15.
5. P(red) = 3/20 + 4/15 = 25/60.
6. Conditional probability = P(A and red)/P(red) = (9/60)/(25/60) = 9/25.

• P1 Establishing $3/5\times 1/4$ or an equivalent valid method.
• P1 Establishing $2/5\times 2/3$ or an equivalent valid method.
• P1 Establishing $(3/20)/(3/20+4/15)$ or an equivalent valid method.
• A1 Correct answer: 9/25

(a) $7/15$

1. Find each mutually exclusive box-and-red route.
   $$2/3\times 3/5+1/3\times 1/5$$
2. Therefore $7/15$.

• M1 Find each mutually exclusive box-and-red route.
• A1 Correct answer: $7/15$

(b) $6/7$

1. Divide the A-and-red joint probability by the total red probability.
   $$(2/3\times 3/5)/(7/15)$$
2. Therefore $6/7$.

• M1 Divide the A-and-red joint probability by the total red probability.
• A1 Correct answer: $6/7$

(a) $\frac{12}{19}$

1. Late and raining.
   $$0.3 \times 0.4 = 0.12$$
2. Late and not raining.
   $$0.7 \times 0.1 = 0.07$$
3. Given late, the total is $0.12 + 0.07 = 0.19$.
   $$P(\text{rain} \mid \text{late}) = \frac{0.12}{0.19} = \frac{12}{19}$$

• P1 $0.3 \times 0.4 = 0.12$.
• P1 $0.7 \times 0.1 = 0.07$ and the total probability of being late, 0.19.
• A1 $\frac{12}{19}$ or equivalent (0.632).

(a) $\frac{19}{68}$

1. Expected frequencies: 20 have the condition; 980 do not.
2. True positives: $0.95 \times 20 = 19$. False positives: $0.05 \times 980 = 49$.
3. Given a positive test:
   $$\frac{19}{19 + 49} = \frac{19}{68}$$

• P1 Finding 20 and 980.
• P1 Both positive groups, 19 and 49.
• P1 Dividing by all positives, 68.
• A1 $\frac{19}{68}$.

(b) Most people who test positive do not have the condition, because the condition is rare.

1. Only 19 of the 68 positives have the condition: about 28%.

• C1 A positive result still means the person is more likely not to have the condition, because false positives outnumber true ones.

(a) 12/35

1. $$0.48+0.44p=0.7$$
2. $$0.48\times 0.5/0.7$$
3. Let P(C works) = p. The probability both A and B work is 0.8 $\times$ 0.6 = 0.48; in that case the alarm operates regardless of C.
4. The probability exactly one of A and B works is 0.8 $\times$ 0.4 + 0.2 $\times$ 0.6 = 0.44. In that case C must work.
5. Thus 0.48 + 0.44p = 0.7, giving p = 0.5.
6. P(all three work) = 0.8 $\times$ 0.6 $\times$ 0.5 = 0.24. Condition on the alarm operating: 0.24/0.7 = 12/35.

• P1 Establishing $0.48+0.44p=0.7$ or an equivalent valid method.
• P1 Establishing $0.48\times 0.5/0.7$ or an equivalent valid method.
• A1 Correct answer: 12/35