mid \text{red first})=\frac{3}{7} $$ 3. If the first counter is blue, there are now 5 red and 2 blue counters left, from 7 counters: $$ P(\text{red second} \mid \text{blue first})=\frac{5}{7}, \quad P(\text{blue second} \mid \text{blue first})=\frac{2}{7} $$ After a counter is taken without replacement, both the number of counters of that colour and the total number of counters change. ## 3. Multiplying along a route A **route** in a tree diagram is one complete path through the tree, such as “red then blue”. For a sequence of events, multiply along the branches: $$ P(A \text{ and } B)=P(A)\times P(B \mid A) $$ A bag has 7 red counters and 5 green counters. Two counters are chosen without replacement. Find the probability that both counters are the same colour. 1. “Same colour” can happen in two separate ways: red then red, or green then green. 2. Multiply along the red-red route: $$ \frac{7}{12}\times\frac{6}{11}=\frac{42}{132} $$ 3. Multiply along the green-green route: $$ \frac{5}{12}\times\frac{4}{11}=\frac{20}{132} $$ 4. Add the two routes because they are separate possibilities: $$ \frac{42}{132}+\frac{20}{132}=\frac{62}{132}=\frac{31}{66} $$ ## 4. Adding separate routes Two outcomes are **mutually exclusive** if they cannot both happen at the same time. For example, “red then red” and “green then green” are mutually exclusive routes. When a question says “one of each” or “different colours”, remember there are usually two possible orders. On Monday, the probability that Sam wears a black tie is 0.55. If he wears black on Monday, the probability he wears black on Tuesday is 0.3. If he does not wear black on Monday, the probability he wears black on Tuesday is 0.8. Find the probability he wears different coloured ties. 1. Different colours means black then not black, or not black then black. 2. Find the useful complements: $$ P(\text{not black on Monday})=1-0.55=0.45 $$ 3. If Monday is black, the probability Tuesday is not black is: $$ 1-0.3=0.7 $$ 4. Multiply along both routes and add: $$ (0.55\times 0.7)+(0.45\times 0.8)=0.385+0.36=0.745 $$ For “one red and one blue”, you need red then blue **and** blue then red, unless the question fixes the order. ## 5. More than two categories If there are three colours or several different number cards, treat each colour or number as its own category. For “same”, add all matching routes. For “second greater than first”, consider what could come second after each possible first card. There are five cards: 1, 1, 2, 2, and 4. Two cards are chosen without replacement. Find the probability that the second card is greater than the first. 1. If the first card is 1, there are 3 greater cards left: 2, 2, and 4. $$ \frac{2}{5}\times\frac{3}{4}=\frac{6}{20} $$ 2. If the first card is 2, there is 1 greater card left: 4. $$ \frac{2}{5}\times\frac{1}{4}=\frac{2}{20} $$ 3. If the first card is 4, there are no greater cards left. $$ \frac{1}{5}\times 0=0 $$ 4. Add the possible routes: $$ \frac{6}{20}+\frac{2}{20}+0=\frac{8}{20}=\frac{2}{5} $$ ## 6. Survey and Venn-style questions A **Venn diagram** sorts people or objects into overlapping groups. An overlap means someone is in both groups. In these questions, first work out how many people are in the group you need. Then use probability without replacement. “Tea and coffee” usually includes people who also like a third drink. “Tea and coffee but not juice” means that exact overlap only.