Probability Rules and Conditional Outcomes for the ESAT

Addition of Probabilities

Events are defined as mutually exclusive if it is impossible for more than one of them to occur at the same time. For two events $A$ and $B$ that are mutually exclusive, the probability of either $A$ or $B$ occurring is the sum of their individual probabilities:

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

If two events are not mutually exclusive, they can happen simultaneously. In this case, there is an overlap that must be accounted for to avoid double counting. The probability is calculated as:

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

Example: Mutually Exclusive Events

Alice has a spinner. The probability of landing on a square number is $\frac{3}{7}$, and the probability of landing on a prime number is $\frac{1}{3}$. What is the probability that the spinner lands on a prime number or a square number?

1. Identify if the events are mutually exclusive. Prime numbers (2, 3, 5, 7...) and square numbers (1, 4, 9, 16...) do not overlap. All square numbers have their square root as a factor, meaning they cannot meet the requirement of having exactly two factors to be prime. Therefore, they are mutually exclusive.
2. Apply the addition rule: $P(\text{prime or square}) = \frac{1}{3} + \frac{3}{7} = \frac{7}{21} + \frac{9}{21} = \frac{16}{21}$.

Example: Testing for Mutual Exclusivity

The probability of event $A$ is $0.32$, and event $B$ is $0.19$. The probability that neither occurs is $0.54$. Are $A$ and $B$ mutually exclusive?

1. Find $P(A \text{ or } B)$: $P(A \text{ or } B) = 1 - P(\text{neither}) = 1 - 0.54 = 0.46$.
2. Use the general addition formula: $0.46 = 0.32 + 0.19 - P(A \text{ and } B)$.
3. Calculate the overlap: $P(A \text{ and } B) = 0.32 + 0.19 - 0.46 = 0.05$.
4. Conclusion: Since $P(A \text{ and } B) \neq 0$, the events are not mutually exclusive.

Multiplication of Probabilities

Two events, $A$ and $B$, are independent if the occurrence of $A$ does not change the probability of $B$ occurring. For independent events, the probability of both occurring is the product of their individual probabilities:

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

Example: Independent Machines

The probability of machine $A$ breaking down is $0.4$, and machine $B$ is $0.25$. One breakdown does not affect the other.

1. Both break down: $P(A \text{ breaks and } B \text{ breaks}) = 0.4 \times 0.25 = 0.1$.
2. Neither breaks down: $P(\text{not } A) \times P(\text{not } B) = (1 - 0.4) \times (1 - 0.25) = 0.6 \times 0.75 = 0.45$.

Tree Diagrams for Independent Events

Tree diagrams help visualise combined events. When events are independent, the probabilities on the second set of branches are identical regardless of the first outcome.

Example: School Sports

A school football team has $P(\text{win}) = 0.6$ and $P(\text{draw}) = 0.15$. A cricket team has $P(\text{win}) = 0.4$ and $P(\text{draw}) = 0.05$. We first find the missing 'lose' probabilities: $P(\text{lose football}) = 1 - 0.6 - 0.15 = 0.25$, and $P(\text{lose cricket}) = 1 - 0.4 - 0.05 = 0.55$.

1. Both teams win: $0.6 \times 0.4 = 0.24$.
2. Both teams lose: $0.25 \times 0.55 = 0.1375$.
3. At least one team wins: This includes all branches where football wins ($0.6$) plus the cases where football does not win but cricket wins ($0.15 \times 0.4 + 0.25 \times 0.4$). Total $= 0.6 + 0.06 + 0.1 = 0.76$.
4. Neither team draws: $(0.6 \times 0.4) + (0.6 \times 0.55) + (0.25 \times 0.4) + (0.25 \times 0.55) = 0.8075$.

Example: Counters with Replacement

A bag has 7 red and 3 blue counters. A counter is picked, returned, then a second is picked. Because the counter is returned, the events are independent.

1. Both red: $\frac{7}{10} \times \frac{7}{10} = \frac{49}{100}$.
2. At least one blue: $1 - P(\text{both red}) = 1 - \frac{49}{100} = \frac{51}{100}$.
3. One of each colour: $(\frac{7}{10} \times \frac{3}{10}) + (\frac{3}{10} \times \frac{7}{10}) = \frac{42}{100} = \frac{21}{50}$.

Tree Diagrams for Dependent Events

When outcomes are dependent, the first event changes the probability of the second. This often occurs when items are not replaced.

Example: Counters without Replacement

Using the same bag (7 red, 3 blue), but the first counter is not replaced.

1. Both red: The first is $\frac{7}{10}$. If red was picked, 6 red remain out of 9 total. Probability $= \frac{7}{10} \times \frac{6}{9} = \frac{42}{90} = \frac{7}{15}$.
2. At least one blue: $1 - P(\text{both red}) = 1 - \frac{7}{15} = \frac{8}{15}$.
3. One of each: $(\frac{7}{10} \times \frac{3}{9}) + (\frac{3}{10} \times \frac{7}{9}) = \frac{42}{90} = \frac{7}{15}$.

Conditional Probability

Conditional probability is the probability of event $B$ occurring known that event $A$ has already occurred. If $A$ and $B$ are independent, $P(B|A) = P(B)$. If they are dependent, these probabilities differ.

Two-Way Tables

Tables are useful for calculating conditional probabilities by restricting the sample space to a specific row or column.

	Production Line	Warehouse	Office	Total
Drive	35	37	21	93
Not Drive	18	20	19	57
Total	53	57	40	150

1. P(Drive given Warehouse): Look only at the warehouse column. $\frac{37}{57}$.
2. P(Office given Drive): Look only at the drive row. $\frac{21}{93} = \frac{7}{31}$.

Venn Diagrams

1. P(studies one language only): $\frac{21 + 9 + 16}{91} = \frac{46}{91}$.
2. P(does not study German): $\frac{21 + 8 + 9 + 6}{91} = \frac{44}{91}$.
3. P(studies French given they study German): Limit to the German circle ($16 + 5 + 19 + 7 = 47$). French students within that group are $19 + 7 = 26$. Probability $= \frac{26}{47}$.

Tree Diagrams for Conditional Probability

To find the probability that the bacterium is present given a positive test result, we use the formula:

$P(\text{present} | \text{positive}) = \frac{P(\text{positive and present})}{P(\text{positive})}$

$P(\text{positive}) = (0.2 \times 0.9) + (0.8 \times 0.07) = 0.18 + 0.056 = 0.236$.

$P(\text{present} | \text{positive}) = \frac{0.18}{0.236} = 0.763$ (to 3 decimal places).