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Statistics and Probability - Probability Theory (Conditional Probability, Bayes' Theorem)

Grade 12IB

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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Conditional Probability: This represents the probability of an event AA occurring given that another event BB has already occurred, denoted as P(A∣B)P(A|B). Visually, this is represented in a Venn diagram by restricting the sample space to the interior of circle BB; we then calculate the proportion of that area occupied by the intersection A∩BA \cap B.

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Independent Events: Two events are independent if the occurrence of one does not affect the probability of the other, meaning P(A∣B)=P(A)P(A|B) = P(A). In a visual tree diagram, this is shown by the second set of branches having the exact same probability values regardless of which outcome occurred in the first set of branches.

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Tree Diagrams: A visual tool used to map out multi-stage probability experiments. Each node splits into branches representing mutually exclusive outcomes. To find the probability of a specific combined outcome, you multiply the probabilities along the path from the root to the leaf; to find the total probability of an outcome that appears on multiple paths, you sum those path products.

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Bayes' Theorem: A formula used to calculate 'inverse' conditional probabilities. It allows us to find P(A∣B)P(A|B) if we already know P(B∣A)P(B|A). Visually, this often involves working 'backward' through a tree diagram to determine the probability that a specific initial branch was taken, given that a certain final result was observed.

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The Law of Total Probability: This law is used to find the overall probability of an event BB by considering all possible ways it can occur through different mutually exclusive events A1,A2,...AnA_1, A_2, ... A_n. Visually, this is the sum of the probabilities of all distinct paths in a tree diagram that end in event BB.

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Mutually Exclusive vs. Independent: Mutually exclusive events cannot occur at the same time (P(A∩B)=0P(A \cap B) = 0). Visually, this is shown as two separate circles in a Venn diagram with no overlap. This is distinct from independence, which refers to the lack of influence one event has on the likelihood of another.

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Complementary Events in Conditional Context: The rule P(A∣B)+P(Aβ€²βˆ£B)=1P(A|B) + P(A'|B) = 1 always holds. Visually, if you are looking strictly inside circle BB on a Venn diagram, any region not belonging to AA must belong to its complement Aβ€²A'.

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Sample Space Reduction: When calculating P(A∣B)P(A|B), the denominator P(B)P(B) acts as the new, reduced sample space. Visually, we ignore every part of the original sample space rectangle that lies outside circle BB and treat BB as the 'new' 100%.

πŸ“Formulae

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

P(A∩B)=P(A∣B)Γ—P(B)P(A \cap B) = P(A|B) \times P(B)

P(A∩B)=P(A)Γ—P(B)Β (ifΒ independent)P(A \cap B) = P(A) \times P(B) \text{ (if independent)}

P(B)=P(B∣A)P(A)+P(B∣Aβ€²)P(Aβ€²)P(B) = P(B|A)P(A) + P(B|A')P(A')

P(A∣B)=P(B∣A)P(A)P(B)P(A|B) = \frac{P(B|A)P(A)}{P(B)}

P(A∣B)=P(B∣A)P(A)P(B∣A)P(A)+P(B∣Aβ€²)P(Aβ€²)P(A|B) = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|A')P(A')}

πŸ’‘Examples

Problem 1:

A box contains 4 red and 6 green markers. Two markers are drawn one after the other without replacement. If the second marker drawn is green, what is the probability that the first marker drawn was also green?

Solution:

Let G1G_1 be the event the first marker is green and G2G_2 be the event the second marker is green. We want to find P(G1∣G2)P(G_1|G_2). \n1. Find P(G1)=610P(G_1) = \frac{6}{10}. \n2. Find P(G2∣G1)P(G_2|G_1): If the first was green, 9 markers remain, 5 of which are green. So, P(G2∣G1)=59P(G_2|G_1) = \frac{5}{9}. \n3. Find P(G1β€²)=410P(G_1') = \frac{4}{10} and P(G2∣G1β€²)P(G_2|G_1'): If the first was red, 9 markers remain, 6 of which are green. So, P(G2∣G1β€²)=69P(G_2|G_1') = \frac{6}{9}. \n4. Total P(G2)=P(G2∣G1)P(G1)+P(G2∣G1β€²)P(G1β€²)=(59)(610)+(69)(410)=3090+2490=5490=0.6P(G_2) = P(G_2|G_1)P(G_1) + P(G_2|G_1')P(G_1') = (\frac{5}{9})(\frac{6}{10}) + (\frac{6}{9})(\frac{4}{10}) = \frac{30}{90} + \frac{24}{90} = \frac{54}{90} = 0.6. \n5. Apply Bayes' Theorem: P(G1∣G2)=P(G2∣G1)P(G1)P(G2)=30/9054/90=3054=59β‰ˆ0.556P(G_1|G_2) = \frac{P(G_2|G_1)P(G_1)}{P(G_2)} = \frac{30/90}{54/90} = \frac{30}{54} = \frac{5}{9} \approx 0.556.

Explanation:

This problem uses Bayes' Theorem. We first calculate the total probability of the evidence (the second marker being green) using the Law of Total Probability, and then find the portion of that probability that came from the 'Green-Green' path.

Problem 2:

Given that P(A)=0.3P(A) = 0.3, P(B)=0.6P(B) = 0.6, and P(AβˆͺB)=0.7P(A \cup B) = 0.7, determine if events AA and BB are independent.

Solution:

  1. Use the Addition Rule to find P(A∩B)P(A \cap B): P(AβˆͺB)=P(A)+P(B)βˆ’P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B). \n2. 0.7=0.3+0.6βˆ’P(A∩B)β€…β€ŠβŸΉβ€…β€ŠP(A∩B)=0.9βˆ’0.7=0.20.7 = 0.3 + 0.6 - P(A \cap B) \implies P(A \cap B) = 0.9 - 0.7 = 0.2. \n3. Check the condition for independence: P(A∩B)=P(A)Γ—P(B)P(A \cap B) = P(A) \times P(B). \n4. P(A)Γ—P(B)=0.3Γ—0.6=0.18P(A) \times P(B) = 0.3 \times 0.6 = 0.18. \n5. Since 0.2β‰ 0.180.2 \neq 0.18, the events are not independent.

Explanation:

To check for independence, we compare the calculated intersection probability with the product of the individual probabilities. If they are not equal, the occurrence of one event affects the probability of the other.