Probability with current chain
$begingroup$
I have this chain.
$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
asked Dec 4 '18 at 20:20
AtstovasAtstovas
1089
$endgroup$
add a comment |
$begingroup$
I have this chain.
$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
asked Dec 4 '18 at 20:20
AtstovasAtstovas
1089
$endgroup$
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
add a comment |
$begingroup$
I have this chain.
$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
asked Dec 4 '18 at 20:20
AtstovasAtstovas
1089
$endgroup$
I have this chain.
$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
probability probability-theory independence
asked Dec 4 '18 at 20:20
AtstovasAtstovas
1089
asked Dec 4 '18 at 20:20
AtstovasAtstovas
1089
asked Dec 4 '18 at 20:20
AtstovasAtstovas
1089
asked Dec 4 '18 at 20:20
AtstovasAtstovas
1089
asked Dec 4 '18 at 20:20
AtstovasAtstovas
1089
1089
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
add a comment |
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
add a comment |
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$begingroup$
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