Probability with current chain












0












$begingroup$


I have this chain.



enter image description here



$A,B,C,D,E $ are switches. They are independent from each other. They can be switched on with probability $p$ or switched off with probability $1-p$. We need to find probability that when current get into the chain it will get out of the chain.



So I was thinking in to ways. Not sure which one is correct if it is so.



First way. We have 4 possibilities to get out from the chain: $AB, AED, CD, CEB$ $$p^2+p^3+p^2+p^2=2p^2(1+p)$$



Second way. $$mathbb{P}big( (Acap B)cup (Acap Ecap D) cup (Ccap D) cup(C cap E cap B) big)$$ but here I have another problem. I know that $mathbb{P}(X cup Y)=mathbb{P}(X)+mathbb{P}(Y)-mathbb{P}(Xcap Y)$ but when I try to apply I always loose something and cannot find right probability....



So which way is correct?










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$endgroup$








  • 1




    $begingroup$
    You can't add probabilities that way...it's possible that multiple pathways are open. Note that if $p=1$ your answer is an unphysical $4$.
    $endgroup$
    – lulu
    Dec 4 '18 at 20:24










  • $begingroup$
    I think the number of combinations blocking current is rather smaller than the number of combinations allowing it. So, I would suggest calculating $P'$, the probability that current can't get out. Then your result $P = 1-P'$.
    $endgroup$
    – bob.sacamento
    Dec 4 '18 at 20:30












  • $begingroup$
    Is it possible to get $$frac{1}{2}p^2+frac{1}{3}p^3$$?
    $endgroup$
    – Atstovas
    Dec 4 '18 at 20:34
















0












$begingroup$


I have this chain.



enter image description here



$A,B,C,D,E $ are switches. They are independent from each other. They can be switched on with probability $p$ or switched off with probability $1-p$. We need to find probability that when current get into the chain it will get out of the chain.



So I was thinking in to ways. Not sure which one is correct if it is so.



First way. We have 4 possibilities to get out from the chain: $AB, AED, CD, CEB$ $$p^2+p^3+p^2+p^2=2p^2(1+p)$$



Second way. $$mathbb{P}big( (Acap B)cup (Acap Ecap D) cup (Ccap D) cup(C cap E cap B) big)$$ but here I have another problem. I know that $mathbb{P}(X cup Y)=mathbb{P}(X)+mathbb{P}(Y)-mathbb{P}(Xcap Y)$ but when I try to apply I always loose something and cannot find right probability....



So which way is correct?










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You can't add probabilities that way...it's possible that multiple pathways are open. Note that if $p=1$ your answer is an unphysical $4$.
    $endgroup$
    – lulu
    Dec 4 '18 at 20:24










  • $begingroup$
    I think the number of combinations blocking current is rather smaller than the number of combinations allowing it. So, I would suggest calculating $P'$, the probability that current can't get out. Then your result $P = 1-P'$.
    $endgroup$
    – bob.sacamento
    Dec 4 '18 at 20:30












  • $begingroup$
    Is it possible to get $$frac{1}{2}p^2+frac{1}{3}p^3$$?
    $endgroup$
    – Atstovas
    Dec 4 '18 at 20:34














0












0








0





$begingroup$


I have this chain.



enter image description here



$A,B,C,D,E $ are switches. They are independent from each other. They can be switched on with probability $p$ or switched off with probability $1-p$. We need to find probability that when current get into the chain it will get out of the chain.



So I was thinking in to ways. Not sure which one is correct if it is so.



First way. We have 4 possibilities to get out from the chain: $AB, AED, CD, CEB$ $$p^2+p^3+p^2+p^2=2p^2(1+p)$$



Second way. $$mathbb{P}big( (Acap B)cup (Acap Ecap D) cup (Ccap D) cup(C cap E cap B) big)$$ but here I have another problem. I know that $mathbb{P}(X cup Y)=mathbb{P}(X)+mathbb{P}(Y)-mathbb{P}(Xcap Y)$ but when I try to apply I always loose something and cannot find right probability....



So which way is correct?










share|cite|improve this question









$endgroup$




I have this chain.



enter image description here



$A,B,C,D,E $ are switches. They are independent from each other. They can be switched on with probability $p$ or switched off with probability $1-p$. We need to find probability that when current get into the chain it will get out of the chain.



So I was thinking in to ways. Not sure which one is correct if it is so.



First way. We have 4 possibilities to get out from the chain: $AB, AED, CD, CEB$ $$p^2+p^3+p^2+p^2=2p^2(1+p)$$



Second way. $$mathbb{P}big( (Acap B)cup (Acap Ecap D) cup (Ccap D) cup(C cap E cap B) big)$$ but here I have another problem. I know that $mathbb{P}(X cup Y)=mathbb{P}(X)+mathbb{P}(Y)-mathbb{P}(Xcap Y)$ but when I try to apply I always loose something and cannot find right probability....



So which way is correct?







probability probability-theory independence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 4 '18 at 20:20









AtstovasAtstovas

1088




1088








  • 1




    $begingroup$
    You can't add probabilities that way...it's possible that multiple pathways are open. Note that if $p=1$ your answer is an unphysical $4$.
    $endgroup$
    – lulu
    Dec 4 '18 at 20:24










  • $begingroup$
    I think the number of combinations blocking current is rather smaller than the number of combinations allowing it. So, I would suggest calculating $P'$, the probability that current can't get out. Then your result $P = 1-P'$.
    $endgroup$
    – bob.sacamento
    Dec 4 '18 at 20:30












  • $begingroup$
    Is it possible to get $$frac{1}{2}p^2+frac{1}{3}p^3$$?
    $endgroup$
    – Atstovas
    Dec 4 '18 at 20:34














  • 1




    $begingroup$
    You can't add probabilities that way...it's possible that multiple pathways are open. Note that if $p=1$ your answer is an unphysical $4$.
    $endgroup$
    – lulu
    Dec 4 '18 at 20:24










  • $begingroup$
    I think the number of combinations blocking current is rather smaller than the number of combinations allowing it. So, I would suggest calculating $P'$, the probability that current can't get out. Then your result $P = 1-P'$.
    $endgroup$
    – bob.sacamento
    Dec 4 '18 at 20:30












  • $begingroup$
    Is it possible to get $$frac{1}{2}p^2+frac{1}{3}p^3$$?
    $endgroup$
    – Atstovas
    Dec 4 '18 at 20:34








1




1




$begingroup$
You can't add probabilities that way...it's possible that multiple pathways are open. Note that if $p=1$ your answer is an unphysical $4$.
$endgroup$
– lulu
Dec 4 '18 at 20:24




$begingroup$
You can't add probabilities that way...it's possible that multiple pathways are open. Note that if $p=1$ your answer is an unphysical $4$.
$endgroup$
– lulu
Dec 4 '18 at 20:24












$begingroup$
I think the number of combinations blocking current is rather smaller than the number of combinations allowing it. So, I would suggest calculating $P'$, the probability that current can't get out. Then your result $P = 1-P'$.
$endgroup$
– bob.sacamento
Dec 4 '18 at 20:30






$begingroup$
I think the number of combinations blocking current is rather smaller than the number of combinations allowing it. So, I would suggest calculating $P'$, the probability that current can't get out. Then your result $P = 1-P'$.
$endgroup$
– bob.sacamento
Dec 4 '18 at 20:30














$begingroup$
Is it possible to get $$frac{1}{2}p^2+frac{1}{3}p^3$$?
$endgroup$
– Atstovas
Dec 4 '18 at 20:34




$begingroup$
Is it possible to get $$frac{1}{2}p^2+frac{1}{3}p^3$$?
$endgroup$
– Atstovas
Dec 4 '18 at 20:34










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