UNIT 1 Conditional Probability

 

Lecture Notes: Conditional Probability

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Definition

Conditional probability measures the likelihood of an event occurring given that another event has already occurred. It helps analyze dependencies between events.

Formula

If $A$ and $B$ are two events, and $P(B) > 0$, the conditional probability of $A$ given $B$ is:

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

Where:

• $P(A|B)$: Probability of $A$ given $B$
• $P(A \cap B)$: Probability of both $A$ and $B$ occurring
• $P(B)$: Probability of $B$ occurring

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Solved Problems

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Problem 1

Question: In a class of 50 students, 30 study Mathematics, and 20 study both Mathematics and Physics. If a student studies Physics, what is the probability that they also study Mathematics?
Solution:

$$P(\text{Mathematics | Physics}) = \frac{P(\text{Mathematics} \cap \text{Physics})}{P(\text{Physics})}$$ $$P(\text{Physics}) = \frac{20}{50} = 0.4,\ P(\text{Mathematics} \cap \text{Physics}) = \frac{20}{50} = 0.4$$ $$P(\text{Mathematics | Physics}) = \frac{0.4}{0.4} = 1$$

Answer: $P(\text{Mathematics | Physics}) = 1$

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Problem 2

Question: A die is rolled. If the outcome is an even number, what is the probability that it is a 4?
Solution:

$$P(4 | \text{Even}) = \frac{P(4 \cap \text{Even})}{P(\text{Even})}$$

• $P(4 \cap \text{Even}) = P(4) = \frac{1}{6}$
• $P(\text{Even}) = \frac{3}{6} = \frac{1}{2}$

$$P(4 | \text{Even}) = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{3}$$

Answer: $P(4 | \text{Even}) = \frac{1}{3}$

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Problem 3

Question: A box contains 5 red, 3 green, and 2 blue balls. If a ball is randomly picked and found to be red, what is the probability that it is one of the 3 red balls numbered 1, 2, or 3?
Solution:

$$P(\text{Numbered Red | Red}) = \frac{P(\text{Numbered Red} \cap \text{Red})}{P(\text{Red})}$$

• $P(\text{Red}) = \frac{5}{10} = 0.5$
• $P(\text{Numbered Red} \cap \text{Red}) = \frac{3}{10} = 0.3$

$$P(\text{Numbered Red | Red}) = \frac{0.3}{0.5} = 0.6$$

Answer: $P(\text{Numbered Red | Red}) = 0.6$

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Problem 4

Question: A student is selected randomly. The probability that they study both Chemistry and Biology is 0.2, and the probability that they study Biology is 0.5. What is the probability that a student studies Chemistry, given that they study Biology?
Solution:

$$P(\text{Chemistry | Biology}) = \frac{P(\text{Chemistry} \cap \text{Biology})}{P(\text{Biology})}$$ $$P(\text{Chemistry | Biology}) = \frac{0.2}{0.5} = 0.4$$

Answer: $P(\text{Chemistry | Biology}) = 0.4$

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Problem 5

Question: In a company, 60% of employees are male, and 70% of males are engineers. What is the probability that a randomly chosen engineer is male?
Solution: Let $M$ = Male, $E$ = Engineer.

$$P(M|E) = \frac{P(M \cap E)}{P(E)}$$

• $P(M) = 0.6, P(E|M) = 0.7$

$$P(M \cap E) = P(M) \cdot P(E|M) = 0.6 \cdot 0.7 = 0.42$$

Assume $P(E) = 0.5$ (given context or additional data).

$$P(M|E) = \frac{0.42}{0.5} = 0.84$$

Answer: $P(M|E) = 0.84$

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Summary

1. Conditional probability measures the dependency of events.
2. Formula: $P(A|B) = \frac{P(A \cap B)}{P(B)}$.
3. Practice problems enhance understanding of real-life applications.