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Law of total probability

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Intros
Lessons
  2. Illustrating the Law of Total Probability for two conditional events
  3. The Law of Total Probability for more than two events
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Examples
Lessons
  1. 2 disjoint mutually exclusive events
    Back Country skiers can be divided into two classes, those with avalanche training and those who have no avalanche training. An individual with avalanche training has a probability of 0.05 of getting in an avalanche, while an individual without avalanche training has a probability of 0.20 of being in an avalanche. If 75% of backcountry skiers have avalanche training and 25% of back country skiers do not, then what is the probability that a randomly selected backcountry skier will be in an avalanche?
    1. 3 or more disjoint mutually exclusive events
      I have 3 bags that each contains 5 marbles.

      Bag 1:

      Bag 2:

      Bag 3:

      2 Green

      4 Green

      5 Green

      3 Red

      1 Red

      0 Red



      I choose one bag at random and draw a marble. What is the probability that I draw a green marble?
      1. Thomas is frequently late to work. If it is sunny he will be late with probability 0.15; if it rains he is late with probability 0.05. And if it snows he is late with probability 0.5 (if he shows up at all). The meteorologist has predicted there is a 50% it will be sunny tomorrow, a 35% chance it will rain and a 15% chance it will snow. What is the probability that Thomas will be late tomorrow?

        Make jokes about taking sick days + make pictures for the sun, rain and snow.
        Topic Notes
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        Recall:
        P(A and B)=P(A)⋅\cdotP(B|A) or equivalently, P(A and B)=P(B)⋅\cdotP(A|B)

        The Law of Total Probability:
        P(A)=P(A and B)+P(A and ~B)=P(B)P(A|B)+P(~B)P(A|~B)

        Or in full generality, if all of B1,B2,...BnB_1, B_2,...B_n include the entire sample space S, and are all pairwise mutually exclusive then:

        P(A)=P(AP(A)=P(A and B1)+P(AB_1)+P(A and B2)+⋯+P(AB_2)+ \cdots +P(A and Bn)B_n)
        =P(B1)P(A∣B1)+P(B2)P(A∣B2)+⋯+P(Bn)P(A∣Bn)=P(B_1)P(A|B_1)+P(B_2)P(A|B_2)+ \cdots + P(B_n)P(A|B_n)