Conditional Probability and Law of Independence

Analysis of Statistics Problem Sheet 4

The problem sheet primarily focuses on Conditional Probability, Independence, and the Law of Total Probability. Below is a summary of key concepts along with brief explanations and small examples.

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Key Concepts and Their Explanations

1. Conditional Probability

Conditional probability measures the probability of an event occurring given that another event has already occurred. It is defined as:

$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$$

where:

• $P(A\mid B)$ is the probability of $A$ given $B$,
• $P(A\cap B)$ is the probability of both $A$ and $B$ happening together,
• $P(B)$ is the probability of $B$.

Example:
A factory has 60% male and 40% female workers. If 30% of male workers smoke and 10% of female workers smoke, what is the probability that a randomly chosen smoker is female?

Using Bayes’ Theorem:

$$P(F\mid S)=\frac{P(S\mid F)P(F)}{P(S)}=\frac{0.1\times 0.4}{(0.1\times 0.4)+(0.3\times 0.6)}=\frac{0.04}{0.22}\approx 0.1818$$

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2. Law of Total Probability

This law allows us to find the probability of an event by considering all possible ways it can occur. It states:

$$P(H)=\sum _{i}P(H\mid A_i)P(A_i)$$

Example:
An engine fails due to:

• Ignition issues ($P(I)=0.5$, service can help $P(H\mid I)=0.6$),
• Fuel supply issues ($P(F)=0.3$, service can help $P(H\mid F)=0.45$),
• Other reasons ($P(O)=0.2$, service can help $P(H\mid O)=0.05$).

Probability that the service helps:

$$P(H)=(0.6\times 0.5)+(0.45\times 0.3)+(0.05\times 0.2)=0.445$$

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3. Inclusion-Exclusion Principle

This principle helps find the probability of the union of multiple events while avoiding overcounting.

For three events $A,B,C$:

$$P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$$

Example:
A store sells apples ($P(A)=0.6$), bananas ($P(B)=0.5$), and oranges ($P(C)=0.4$). If:

• $P(A\cap B)=0.3$,
• $P(A\cap C)=0.2$,
• $P(B\cap C)=0.25$,
• $P(A\cap B\cap C)=0.1$,

then:

$$P(A\cup B\cup C)=0.6+0.5+0.4-0.3-0.2-0.25+0.1=0.85$$

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4. Independent Events

Two events $A$ and $B$ are independent if:

$$P(A\cap B)=P(A)\times P(B)$$

Example:
Rolling a die ($P(\text{6})=\frac{1}{6}$) and flipping a coin ($P(\text{Heads})=\frac{1}{2}$) are independent.

$$P(\text{6}\cap \text{Heads})=\frac{1}{6}\times \frac{1}{2}=\frac{1}{12}$$

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5. Bayesian Probability

Bayes’ Theorem calculates the probability of an event given prior knowledge:

$$P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B)}$$

Example:
A test detects a rare disease ($P(D)=0.0001$) with 99% accuracy for diseased people ($P(\text{Pos}\mid D)=0.99$) and 95% accuracy for healthy people ($P(\text{Neg}\mid \neg D)=0.95$).

Probability of having the disease given a positive test result:

$$P(D\mid \text{Pos})=\frac{0.99\times 0.0001}{(0.99\times 0.0001)+(0.05\times 0.9999)}\approx 0.002$$