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Compound Probability

Understanding Compound Probability

Compound probability involves finding the likelihood of two or more events occurring. To solve these problems, we use two main rules: the Multiplication Rule for "AND" scenarios and the Addition Rule for "OR" scenarios.

The Multiplication Rule (P(AB)P(A \cap B))

The multiplication rule is used when you want to find the probability of event A and event B happening.

Independent vs. Dependent Events

  • Independent Events: The outcome of the first event does not affect the second. Formula: P(A and B)=P(A)×P(B)P(A \text{ and } B) = P(A) \times P(B)
  • Dependent Events: The outcome of the first event changes the probability of the second (e.g., drawing without replacement). Formula: P(A and B)=P(A)×P(BA)P(A \text{ and } B) = P(A) \times P(B|A) (Where P(BA)P(B|A) is the probability of B given that A has occurred).

Example: Drawing without replacement A bag has 5 red and 3 blue balls. Two are drawn without replacement. Find P(both red)P(\text{both red}).

  1. The probability of drawing a red ball first is P(1st red)=58P(\text{1st red}) = \frac{5}{8}.
  2. Since the ball is not replaced, there are now 4 red balls and 7 total balls left. So, P(2nd red1st red)=47P(\text{2nd red} | \text{1st red}) = \frac{4}{7}.
  3. Multiply them together: P(both red)=58×47=2056=514P(\text{both red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}

The Addition Rule (P(AB)P(A \cup B))

The addition rule is used when you want to find the probability of event A or event B happening (or both).

Formula: P(AB)=P(A)+P(B)P(AB)P(A \cup B) = P(A) + P(B) - P(A \cap B)

We subtract P(AB)P(A \cap B) (the intersection) because simply adding P(A)P(A) and P(B)P(B) double-counts the scenario where both events happen at the same time.

Example: Finding the Union Given P(A)=0.4P(A) = 0.4, P(B)=0.5P(B) = 0.5, and P(AB)=0.15P(A \cap B) = 0.15. Find P(AB)P(A \cup B).

P(AB)=0.4+0.50.15=0.75P(A \cup B) = 0.4 + 0.5 - 0.15 = 0.75

Testing for Independence

You can mathematically prove whether two events are independent. Events AA and BB are strictly independent if and only if their intersection equals the product of their individual probabilities:

P(AB)=P(A)×P(B)P(A \cap B) = P(A) \times P(B)

Example: Are A and B independent? Using the values from the previous example: P(A)=0.4P(A) = 0.4, P(B)=0.5P(B) = 0.5, and P(AB)=0.15P(A \cap B) = 0.15.

  1. Calculate P(A)×P(B)P(A) \times P(B): 0.4×0.5=0.200.4 \times 0.5 = 0.20
  2. Compare to P(AB)P(A \cap B): 0.200.150.20 \neq 0.15

Because the products are not equal, events AA and BB are dependent.