As promised last time, I am going to cover Baye's Theorem.
If Tree diagram is the common name for Bayes Theorem. Recall that conditional probability is given by $P(A \mid B) = \frac{P(A \wedge B)}{P(B)}$. For tree diagrams,
let's say that we have events A, B1, B2, B3, … (the reason we have
multiple B's is because they all are within the same family of
events) such that the events in the family of B are mutually
exclusive and the sum of the probabilities of the events in the
family of B are equal to 1. Then we have $$P(B_i \mid A)= \frac{P(B_i)*P(A \mid B_i)}{\sum_{m=1}^{n}[P(B_m)*P(A \mid B_m)]}$$ What this means is reliant on the tree diagram.
If J has a probability of 100%, and P(C) and P(D) are not 0, then when we are trying to find the probability of any of the B's being true given that A is true, we have to set the probability of A to be the entirety of the probability. Basically, excluding the probabilities of C and D, because those would be guaranteed as not happening if we have B given A. To put it another way, saying "B given A" sets the probability of A occurring equal to 100%.
K. "Alan" Eister has his bacholers of Science in Chemistry. He is also a tutor for Varsity Tutors. If you feel you need any further help on any of the topics covered, you can sign up for tutoring sessions here.
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| If we are only looking at the sub-items of A, this is what the tree diagram would look like. |
If J has a probability of 100%, and P(C) and P(D) are not 0, then when we are trying to find the probability of any of the B's being true given that A is true, we have to set the probability of A to be the entirety of the probability. Basically, excluding the probabilities of C and D, because those would be guaranteed as not happening if we have B given A. To put it another way, saying "B given A" sets the probability of A occurring equal to 100%.
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| Fully Realized Bayes Theorem Tree Diagrams. |
Here, the sum of all probabilities on Level 1 is equal to unity (is equal to 1). The sum of all level 2 events is also equal to 1. The same applies to level 3. This can be stated as "For any given level, $\sum_{m=1}^{n}P_{L}(M_{m})=1$ always applies. Notice that this is just individual probabilities, not probabilities of the level given it's corresponding entry in the previous level. This means that all of the probabilities in the level have to add up to 1, independent of which entry in the previous level was given. If we have the statement of "Probability of one thing given another thing", than each branch adds up to one. This comes from what I've mentioned before about the fact that $P(A \mid B)=\frac{P(A \wedge B)}{P(A)}$. This sets the probability of the parent group B equal to one, and the probability of all other objects in the same level equal to zero.
That's the concept of Baye's Theorem. If you have any questions, please leave it in the comments. Next time, I'll cover some common plots of physical data. Until then, stay curious.
That's the concept of Baye's Theorem. If you have any questions, please leave it in the comments. Next time, I'll cover some common plots of physical data. Until then, stay curious.
K. "Alan" Eister has his bacholers of Science in Chemistry. He is also a tutor for Varsity Tutors. If you feel you need any further help on any of the topics covered, you can sign up for tutoring sessions here.
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