MHB A question about the law of total probability

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The discussion focuses on calculating the probability that goat cheese and parmesan cheese are on opposite sides of a randomly placed pickle between n different cheeses. Participants suggest using the law of total probability and defining events based on the pickle's position. One approach proposed is to consider the two cheeses as a single unit to simplify counting the arrangements. The challenge lies in determining the total arrangements and the specific configurations where the two cheeses are separated by the pickle. This problem illustrates the application of probability theory in combinatorial scenarios.
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Dan put n different kinds of cheese in a row randomly, so that between two kinds of cheese there is a space. Then he puts a pickle on one of the n-1 spaces between the cheeses randomly. What is the probability of goat cheese and parmesan cheese (2 from n kinds of cheese) to be in the different sides of the pickle?

I know that I need to use law of total probability and define Ai as the pickle is found in the space number I from n-1 spaces, but I am stuck.
 
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lola19991 said:
Dan put n different kinds of cheese in a row randomly, so that between two kinds of cheese there is a space. Then he puts a pickle on one of the n-1 spaces between the cheeses randomly. What is the probability of goat cheese and parmesan cheese (2 from n kinds of cheese) to be in the different sides of the pickle?

I know that I need to use law of total probability and define Ai as the pickle is found in the space number I from n-1 spaces, but I am stuck.

Don't we first need the probability that GOAT and PARM end up together? Perhaps what may seem like an odd approach, tape those two together and see if it is easier to count the possibilities.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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