Probability of current flowing through circuits

[Jameson]

I have two solutions to the this problem that I've found but they seem strange compared to each other.

Solution I
All possible paths that work must pass through circuit 5, so we can factor out that probability and tack it onto the end because they are all independent.

$$P(A \rightarrow D)=P[(1 \cap 2) \cup (3 \cap 4) \cup (1 \cap 4) \cup (3 \cap 2) \cap 5] $$

The above just shows that it can go through the path 1,2,5 or 3,4,5 or 1,4,5 or 3,2,5. It has to be at least one of them but it can be more. Now using independence and complements this simplifies to:

$$P(A \rightarrow D)=\left(1 - P \left[ (1 \cap 2)' \cap (3 \cap 4)' \cap (1 \cap 4)' \cap (3 \cap 2)' \right]
\right)*p_5=\boxed{\left[1 - (1-p_1p_2)(1-p_3p_4)(1-p_1p_4)(1-p_3p_2)
\right]p_5}$$

Solution II
Using a similar argument but instead of listing the paths I will list the stages. It has to pass through 1 or 3 initially, then 2 or 4 and finally 5.

$$P(A \rightarrow D)=P[(1 \cup 3) \cap (2 \cup 4) \cap 5]$$.

Now using independence and the inclusion-exclusion principle I get the following answer:

$$P(A \rightarrow D) = \boxed{(p_1+p_3-p_1p_3)(p_2+p_4-p_2p_4)p_5}$$

So my main question is do you agree with both of these answers? Any comments? :)

 

[I like Serena]

Hi Jameson!

The idea of solution I is fine, but it does not work like that because the probabilities are not independent.
Btw, please add couple of parentheses in the first line - it hurts my eyes to see something that is incorrect.

Solution II is fine.

 

[Jameson]

Hi Jameson!

The idea of solution I is fine, but it does not work like that because the probabilities are not independent.
Btw, please add couple of parentheses in the first line - it hurts my eyes to see something that is incorrect.

Solution II is fine.

Hi I like Serena! :)

The probabilities are independent actually. That's the only reason I can factor them at all like I did. Is there some way I did it incorrectly?

Where should I add parentheses? I admit it doesn't look the best but I tried to format it as nicely as I could.

 

[I like Serena]

Hi I like Serena! :)

The probabilities are independent actually. That's the only reason I can factor them at all like I did. Is there some way I did it incorrectly?

Where should I add parentheses? I admit it doesn't look the best but I tried to format it as nicely as I could.

It should be:
$$P(A \rightarrow D)=P\left[\left(\phantom{\frac{}{}}(1 \cap 2) \cup (3 \cap 4) \cup (1 \cap 4) \cup (3 \cap 2)\right) \cap 5\right]$$
In the following calculation the probabilities are not independent, because 2 appears in both of them.
$$P \left[ (1 \cap 2)' \cap (3 \cap 2)' \right] \ne P \left[ (1 \cap 2)' \right] P\left[ (3 \cap 2)' \right]$$

 

[Jameson]

So you're saying the LHS actually simplifies to $$(1-p_1)(1-p_2)(1-p_3)$$ but the RHS will be $$(1-p_1)(1-p_2)^2(1-p_3)$$. If so then that makes sense.

So assuming this idea is correct, $$P(A \rightarrow D)=P\left[\left(\phantom{\frac{}{}}(1 \cap 2) \cup (3 \cap 4) \cup (1 \cap 4) \cup (3 \cap 2)\right) \cap 5\right]$$, then how would you proceed to simplify it?

 

[I like Serena]

So you're saying the LHS actually simplifies to $$(1-p_1)(1-p_2)(1-p_3)$$ but the RHS will be $$(1-p_1)(1-p_2)^2(1-p_3)$$. If so then that makes sense.

Not quite. You are forgetting the negations.

So assuming this idea is correct, $$P(A \rightarrow D)=P\left[\left(\phantom{\frac{}{}}(1 \cap 2) \cup (3 \cap 4) \cup (1 \cap 4) \cup (3 \cap 2)\right) \cap 5\right]$$, then how would you proceed to simplify it?

I wouldn't.
It becomes way to complex.
I suspect it would be best to reduce the set expression to one without duplications.
We can assume that you'll simply get the other expression.

 

[Jameson]

Ok, I'll just forget that one. Solution II is very short and easy to do so I'll stick with that method. Luckily that's the method I chose to do today on my midterm, haha. :p

Thanks again for all your help.