Combined Probability

Basic Probability Concepts

An event is something that can happen in a probability experiment, and an outcome is a single possible result of that event. For example, when rolling a die, an outcome could be rolling a 3, and the event could be "rolling an even number".

Probability values always lie between 0 and 1 inclusive:

• 0 means the event cannot happen (impossible event)
• 1 means the event is certain to happen
• Values between 0 and 1 show how likely the event is

Mutually exclusive events are events that cannot happen at the same time. For example, when flipping a coin, getting heads and tails are mutually exclusive because both cannot occur on the same flip.

For instance, the probability of rolling a 4 or 5 on a die is $\frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$.

Combined Probability Rules

Addition Rule for 'Or' Events

When finding the probability of one event or another event happening, use the addition rule:

$$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$$

If the events are mutually exclusive, then they cannot both happen, so:

$$P(A \text{ or } B) = P(A) + P(B)$$

For example, when rolling a die, the probability of rolling a 2 or a 5 is:

$$P(2 \text{ or } 5) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$$

Multiplication Rule for Independent Events

When two events are independent (the outcome of one does not affect the other), the probability of both events happening is the product of their probabilities:

$$P(A \text{ and } B) = P(A) \times P(B)$$

For example, tossing two coins, the probability of getting heads on both is:

$$P(\text{Heads on first coin and Heads on second coin}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$

Complementary Events

The complement of an event $A$ is the event that $A$ does not happen, written as $A'$. The probabilities of an event and its complement always add up to 1:

$$P(A) + P(A') = 1 \quad \Rightarrow \quad P(A') = 1 - P(A)$$

For example, if the probability of raining tomorrow is 0.3, then the probability of not raining is:

$$P(\text{not rain}) = 1 - 0.3 = 0.7$$

Using Probability Diagrams

Tree Diagrams for Sequences

Tree diagrams are useful for showing all possible outcomes of a sequence of events. Each branch represents an outcome with its probability. Multiply probabilities along branches to find combined probabilities.

For example, when tossing two coins, the tree diagram shows four possible outcomes: HH, HT, TH, TT, each with probability $\frac{1}{4}$.

To find the probability of getting exactly one head, add the probabilities of HT and TH:

$$P(\text{exactly one head}) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$

Sample Space Diagrams

Sample space diagrams list all possible outcomes of an experiment, often in a grid or table format. They help visualise combined events and calculate probabilities by counting outcomes.

For example, when rolling two dice, the sample space diagram shows 36 possible outcomes (6 $\times$ 6). To find the probability of the sum being 7, count all pairs that add to 7 (there are 6), so:

$$P(\text{sum of 7}) = \frac{6}{36} = \frac{1}{6}$$

Calculating Combined Probabilities

Use tree diagrams or sample space diagrams to calculate combined probabilities by:

• Multiplying probabilities along branches for 'and' events
• Adding probabilities of mutually exclusive outcomes for 'or' events

Example: A bag contains 3 red and 2 blue balls. Two balls are drawn one after the other with replacement. Find the probability both balls are red.

Since the draws are independent (replacement),

$$P(\text{red on first draw}) = \frac{3}{5} \quad \text{and} \quad P(\text{red on second draw}) = \frac{3}{5}$$

So combined probability is:

$$\frac{3}{5} \times \frac{3}{5} = \frac{9}{25}$$

This can be shown clearly on a tree diagram with branches labelled with probabilities.

For events without replacement, probabilities change after the first event because the total number of outcomes changes.

For example, if two balls are drawn without replacement, the probability both are red is:

$$P(\text{red first}) = \frac{3}{5}, \quad P(\text{red second} | \text{red first}) = \frac{2}{4}$$

So combined probability:

$$\frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10}$$