Mixing
Intersection and conditioned set
0
I don't understand why if it's true that $P(Acap B|C)=P(A|C)cdot P(B|C)$ (formula, moreover, used in a number of exercises), in this case he writes:
probability statistics elementary-set-theory boolean-algebra
asked Nov 21 '18 at 10:19
Marco Pittella
1258
add a comment |
0
I don't understand why if it's true that $P(Acap B|C)=P(A|C)cdot P(B|C)$ (formula, moreover, used in a number of exercises), in this case he writes:
probability statistics elementary-set-theory boolean-algebra
asked Nov 21 '18 at 10:19
Marco Pittella
1258
1
Actually, I do not believe that $P(Acap B|C)=P(A|C)cdot P(B|C)$ was used in the calculation. It seems more to be something like $P(Acap B|C)=P(A|B)cdot P(B|C)$. This is not true in general, there has to be some assumption on $A,B,C$. What is true however is that $P(Acap B|C)=P(A|B cap C)cdot P(B|C)$.
– Matthias Klupsch
Nov 21 '18 at 10:28
@MatthiasKlupsch Thanks for your answer. Well, you're welcome to watch the solution that i post below.
– Marco Pittella
Nov 21 '18 at 10:33
@MatthiasKlupsch I'm sorry for my reponse comment, i don't speak english so well. In any case, the photo up here is referring to the following exercise. You have 5 coins and $p_i$ is the probability that the $i$ coin gives Head (T), where $p_1=0,p_2=0,25,p_3=0,5,p_4=0,75,p_5=1$. One coin is extrracted and launched. You obtain Head (T). If the same $i$ coin launched again, what is the probability to obtain another Head (T)?
– Marco Pittella
Nov 21 '18 at 10:48
add a comment |
0
0
0
1
I don't understand why if it's true that $P(Acap B|C)=P(A|C)cdot P(B|C)$ (formula, moreover, used in a number of exercises), in this case he writes:
probability statistics elementary-set-theory boolean-algebra
asked Nov 21 '18 at 10:19
Marco Pittella
1258
I don't understand why if it's true that $P(Acap B|C)=P(A|C)cdot P(B|C)$ (formula, moreover, used in a number of exercises), in this case he writes:
probability statistics elementary-set-theory boolean-algebra
probability statistics elementary-set-theory boolean-algebra
asked Nov 21 '18 at 10:19
Marco Pittella
1258
asked Nov 21 '18 at 10:19
Marco Pittella
1258
asked Nov 21 '18 at 10:19
Marco Pittella
1258
asked Nov 21 '18 at 10:19
Marco Pittella
1258
asked Nov 21 '18 at 10:19
Marco Pittella
1258
1258
1
Actually, I do not believe that $P(Acap B|C)=P(A|C)cdot P(B|C)$ was used in the calculation. It seems more to be something like $P(Acap B|C)=P(A|B)cdot P(B|C)$. This is not true in general, there has to be some assumption on $A,B,C$. What is true however is that $P(Acap B|C)=P(A|B cap C)cdot P(B|C)$.
– Matthias Klupsch
Nov 21 '18 at 10:28
@MatthiasKlupsch Thanks for your answer. Well, you're welcome to watch the solution that i post below.
– Marco Pittella
Nov 21 '18 at 10:33
@MatthiasKlupsch I'm sorry for my reponse comment, i don't speak english so well. In any case, the photo up here is referring to the following exercise. You have 5 coins and $p_i$ is the probability that the $i$ coin gives Head (T), where $p_1=0,p_2=0,25,p_3=0,5,p_4=0,75,p_5=1$. One coin is extrracted and launched. You obtain Head (T). If the same $i$ coin launched again, what is the probability to obtain another Head (T)?
– Marco Pittella
Nov 21 '18 at 10:48
add a comment |
1
Actually, I do not believe that $P(Acap B|C)=P(A|C)cdot P(B|C)$ was used in the calculation. It seems more to be something like $P(Acap B|C)=P(A|B)cdot P(B|C)$. This is not true in general, there has to be some assumption on $A,B,C$. What is true however is that $P(Acap B|C)=P(A|B cap C)cdot P(B|C)$.
– Matthias Klupsch
Nov 21 '18 at 10:28
@MatthiasKlupsch Thanks for your answer. Well, you're welcome to watch the solution that i post below.
– Marco Pittella
Nov 21 '18 at 10:33
@MatthiasKlupsch I'm sorry for my reponse comment, i don't speak english so well. In any case, the photo up here is referring to the following exercise. You have 5 coins and $p_i$ is the probability that the $i$ coin gives Head (T), where $p_1=0,p_2=0,25,p_3=0,5,p_4=0,75,p_5=1$. One coin is extrracted and launched. You obtain Head (T). If the same $i$ coin launched again, what is the probability to obtain another Head (T)?
– Marco Pittella
Nov 21 '18 at 10:48
1
1
Actually, I do not believe that $P(Acap B|C)=P(A|C)cdot P(B|C)$ was used in the calculation. It seems more to be something like $P(Acap B|C)=P(A|B)cdot P(B|C)$. This is not true in general, there has to be some assumption on $A,B,C$. What is true however is that $P(Acap B|C)=P(A|B cap C)cdot P(B|C)$.
– Matthias Klupsch
Nov 21 '18 at 10:28
Actually, I do not believe that $P(Acap B|C)=P(A|C)cdot P(B|C)$ was used in the calculation. It seems more to be something like $P(Acap B|C)=P(A|B)cdot P(B|C)$. This is not true in general, there has to be some assumption on $A,B,C$. What is true however is that $P(Acap B|C)=P(A|B cap C)cdot P(B|C)$.
– Matthias Klupsch
Nov 21 '18 at 10:28
@MatthiasKlupsch Thanks for your answer. Well, you're welcome to watch the solution that i post below.
– Marco Pittella
Nov 21 '18 at 10:33
@MatthiasKlupsch Thanks for your answer. Well, you're welcome to watch the solution that i post below.
– Marco Pittella
Nov 21 '18 at 10:33
@MatthiasKlupsch I'm sorry for my reponse comment, i don't speak english so well. In any case, the photo up here is referring to the following exercise. You have 5 coins and $p_i$ is the probability that the $i$ coin gives Head (T), where $p_1=0,p_2=0,25,p_3=0,5,p_4=0,75,p_5=1$. One coin is extrracted and launched. You obtain Head (T). If the same $i$ coin launched again, what is the probability to obtain another Head (T)?
– Marco Pittella
Nov 21 '18 at 10:48
@MatthiasKlupsch I'm sorry for my reponse comment, i don't speak english so well. In any case, the photo up here is referring to the following exercise. You have 5 coins and $p_i$ is the probability that the $i$ coin gives Head (T), where $p_1=0,p_2=0,25,p_3=0,5,p_4=0,75,p_5=1$. One coin is extrracted and launched. You obtain Head (T). If the same $i$ coin launched again, what is the probability to obtain another Head (T)?
– Marco Pittella
Nov 21 '18 at 10:48
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Example like the following: 
Thanks again.
answered Nov 21 '18 at 10:34
Marco Pittella
1258
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Example like the following:
Thanks again.
answered Nov 21 '18 at 10:34
Marco Pittella
1258
add a comment |
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Example like the following:
Thanks again.
answered Nov 21 '18 at 10:34
Marco Pittella
1258
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Example like the following:
Thanks again.
answered Nov 21 '18 at 10:34
Marco Pittella
1258
Example like the following:
Thanks again.
answered Nov 21 '18 at 10:34
Marco Pittella
1258
answered Nov 21 '18 at 10:34
Marco Pittella
1258
answered Nov 21 '18 at 10:34
Marco Pittella
1258
answered Nov 21 '18 at 10:34
Marco Pittella
1258
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1
Actually, I do not believe that $P(Acap B|C)=P(A|C)cdot P(B|C)$ was used in the calculation. It seems more to be something like $P(Acap B|C)=P(A|B)cdot P(B|C)$. This is not true in general, there has to be some assumption on $A,B,C$. What is true however is that $P(Acap B|C)=P(A|B cap C)cdot P(B|C)$.
– Matthias Klupsch
Nov 21 '18 at 10:28
@MatthiasKlupsch Thanks for your answer. Well, you're welcome to watch the solution that i post below.
– Marco Pittella
Nov 21 '18 at 10:33
@MatthiasKlupsch I'm sorry for my reponse comment, i don't speak english so well. In any case, the photo up here is referring to the following exercise. You have 5 coins and $p_i$ is the probability that the $i$ coin gives Head (T), where $p_1=0,p_2=0,25,p_3=0,5,p_4=0,75,p_5=1$. One coin is extrracted and launched. You obtain Head (T). If the same $i$ coin launched again, what is the probability to obtain another Head (T)?
– Marco Pittella
Nov 21 '18 at 10:48