Compound Probability Math Example 4

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Example 4

hard
Using the law of total probability: P(B)=P(B∣A)P(A)+P(B∣Ac)P(Ac)P(B) = P(B|A)P(A) + P(B|A^c)P(A^c), find P(B)P(B) given P(A)=0.4P(A)=0.4, P(B∣A)=0.7P(B|A)=0.7, P(B∣Ac)=0.3P(B|A^c)=0.3.

Solution

  1. 1
    P(Ac)=1−0.4=0.6P(A^c) = 1 - 0.4 = 0.6
  2. 2
    P(B)=P(B∣A)P(A)+P(B∣Ac)P(Ac)=0.7(0.4)+0.3(0.6)=0.28+0.18=0.46P(B) = P(B|A)P(A) + P(B|A^c)P(A^c) = 0.7(0.4) + 0.3(0.6) = 0.28 + 0.18 = 0.46

Answer

P(B)=0.7(0.4)+0.3(0.6)=0.46P(B) = 0.7(0.4) + 0.3(0.6) = 0.46
The law of total probability decomposes P(B) by conditioning on whether A occurs. It weights conditional probabilities by the probability of each conditioning event. This is the foundation for Bayes' theorem and is useful when P(B) is unknown but conditional probabilities are available.

About Compound Probability

The probability of two or more events occurring together (P(A and B)P(A \text{ and } B)) or at least one occurring (P(A or B)P(A \text{ or } B)), accounting for whether the events are independent or dependent.

Learn more about Compound Probability →

More Compound Probability Examples

Example 1 medium

Events A and B: [formula], [formula], [formula]. Find (a) [formula], (b) [formula], and verify wheth

Example 2 hard

A card is drawn from a standard deck. Event A: card is red. Event B: card is a face card (J, Q, K).

Example 3 easy

[formula], [formula], and A and B are mutually exclusive. Find [formula].

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