# Law of total probability

##### Intros

###### Lessons

##### Examples

###### Lessons

**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?**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?- 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.

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###### Topic Notes

Recall:

P(

Or in full generality, if all of $B_1, B_2,...B_n$ include the entire sample space S, and are all pairwise mutually exclusive then:

$P(A)=P(A$ and $B_1)+P(A$ and $B_2)+ \cdots +P(A$ and $B_n)$

$=P(B_1)P(A|B_1)+P(B_2)P(A|B_2)+ \cdots + P(B_n)P(A|B_n)$

P(

**and***A***)=P(***B***)$\cdot$P(***A***|***B***) or equivalently, P(***A***and***A***)=P(***B***)$\cdot$P(***B***|***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 $B_1, B_2,...B_n$ include the entire sample space S, and are all pairwise mutually exclusive then:

$P(A)=P(A$ and $B_1)+P(A$ and $B_2)+ \cdots +P(A$ and $B_n)$

$=P(B_1)P(A|B_1)+P(B_2)P(A|B_2)+ \cdots + P(B_n)P(A|B_n)$

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