Intersection and conditioned set
up vote
0
down vote
favorite
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 at 10:19
Marco Pittella
977
add a comment |
up vote
0
down vote
favorite
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 at 10:19
Marco Pittella
977
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 at 10:28
@MatthiasKlupsch Thanks for your answer. Well, you're welcome to watch the solution that i post below.
– Marco Pittella
Nov 21 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 at 10:48
add a comment |
up vote
0
down vote
favorite
1
up vote
0
down vote
favorite
1
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 at 10:19
Marco Pittella
977
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 at 10:19
Marco Pittella
977
asked Nov 21 at 10:19
Marco Pittella
977
asked Nov 21 at 10:19
Marco Pittella
977
asked Nov 21 at 10:19
Marco Pittella
977
asked Nov 21 at 10:19
Marco Pittella
977
977
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 at 10:28
@MatthiasKlupsch Thanks for your answer. Well, you're welcome to watch the solution that i post below.
– Marco Pittella
Nov 21 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 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 at 10:28
@MatthiasKlupsch Thanks for your answer. Well, you're welcome to watch the solution that i post below.
– Marco Pittella
Nov 21 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 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 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 at 10:28
@MatthiasKlupsch Thanks for your answer. Well, you're welcome to watch the solution that i post below.
– Marco Pittella
Nov 21 at 10:33
@MatthiasKlupsch Thanks for your answer. Well, you're welcome to watch the solution that i post below.
– Marco Pittella
Nov 21 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 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 at 10:48
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Example like the following: 
Thanks again.
answered Nov 21 at 10:34
Marco Pittella
977
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Example like the following:
Thanks again.
answered Nov 21 at 10:34
Marco Pittella
977
add a comment |
up vote
0
down vote
Example like the following:
Thanks again.
answered Nov 21 at 10:34
Marco Pittella
977
add a comment |
up vote
0
down vote
up vote
0
down vote
Example like the following:
Thanks again.
answered Nov 21 at 10:34
Marco Pittella
977
Example like the following:
Thanks again.
answered Nov 21 at 10:34
Marco Pittella
977
answered Nov 21 at 10:34
Marco Pittella
977
answered Nov 21 at 10:34
Marco Pittella
977
answered Nov 21 at 10:34
Marco Pittella
977
977
add a comment |
add a comment |
draft saved
draft discarded
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007529%2fintersection-and-conditioned-set%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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 at 10:28
@MatthiasKlupsch Thanks for your answer. Well, you're welcome to watch the solution that i post below.
– Marco Pittella
Nov 21 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 at 10:48