Is $P(E/bar F) + P(bar E/ bar F) = 1$?
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1
down vote
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Is $P(E/bar F) + P(bar E/ bar F) = 1$?
I tried using venn diagram and here's what I get:
The red region shows $P(E/ bar F)$ and the blue region shows $P(bar E/ bar F)$
Clearly it is not $1$ but my book says it is. Where have I misunderstood?
probability
asked Nov 16 at 6:28
Abcd
2,89421130
add a comment |
up vote
1
down vote
favorite
Is $P(E/bar F) + P(bar E/ bar F) = 1$?
I tried using venn diagram and here's what I get:
The red region shows $P(E/ bar F)$ and the blue region shows $P(bar E/ bar F)$
Clearly it is not $1$ but my book says it is. Where have I misunderstood?
probability
asked Nov 16 at 6:28
Abcd
2,89421130
You probably confuse the relative complement and the conditional probability.
– Hanul Jeon
Nov 16 at 6:30
You are confusing $/$ , that is conditional probability, with $cap$, the intersection. Conditional probability $P(E / F)$ requires that you take the ratio $frac{P(E cap F)}{P(F)}$, so it is not the probability of some event in the space, but rather a ratio of two probabilities, which can't be represented on the Venn diagram.
– астон вілла олоф мэллбэрг
Nov 16 at 6:30
Is $P(E/F)$ a conditional probability. If so, then recall the definition of conditional probability.
– Lord Shark the Unknown
Nov 16 at 6:31
Is it not possible to sketch conditional probability on venn diagram @LordSharktheUnknown
– Abcd
Nov 16 at 6:33
@Abcd You can sketch it. Just cut out the circle for $F$. Now, this is your new sample space.
– trancelocation
Nov 16 at 7:17
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Is $P(E/bar F) + P(bar E/ bar F) = 1$?
I tried using venn diagram and here's what I get:
The red region shows $P(E/ bar F)$ and the blue region shows $P(bar E/ bar F)$
Clearly it is not $1$ but my book says it is. Where have I misunderstood?
probability
asked Nov 16 at 6:28
Abcd
2,89421130
Is $P(E/bar F) + P(bar E/ bar F) = 1$?
I tried using venn diagram and here's what I get:
The red region shows $P(E/ bar F)$ and the blue region shows $P(bar E/ bar F)$
Clearly it is not $1$ but my book says it is. Where have I misunderstood?
probability
probability
asked Nov 16 at 6:28
Abcd
2,89421130
asked Nov 16 at 6:28
Abcd
2,89421130
asked Nov 16 at 6:28
Abcd
2,89421130
asked Nov 16 at 6:28
Abcd
2,89421130
asked Nov 16 at 6:28
Abcd
2,89421130
2,89421130
You probably confuse the relative complement and the conditional probability.
– Hanul Jeon
Nov 16 at 6:30
You are confusing $/$ , that is conditional probability, with $cap$, the intersection. Conditional probability $P(E / F)$ requires that you take the ratio $frac{P(E cap F)}{P(F)}$, so it is not the probability of some event in the space, but rather a ratio of two probabilities, which can't be represented on the Venn diagram.
– астон вілла олоф мэллбэрг
Nov 16 at 6:30
Is $P(E/F)$ a conditional probability. If so, then recall the definition of conditional probability.
– Lord Shark the Unknown
Nov 16 at 6:31
Is it not possible to sketch conditional probability on venn diagram @LordSharktheUnknown
– Abcd
Nov 16 at 6:33
@Abcd You can sketch it. Just cut out the circle for $F$. Now, this is your new sample space.
– trancelocation
Nov 16 at 7:17
add a comment |
You probably confuse the relative complement and the conditional probability.
– Hanul Jeon
Nov 16 at 6:30
You are confusing $/$ , that is conditional probability, with $cap$, the intersection. Conditional probability $P(E / F)$ requires that you take the ratio $frac{P(E cap F)}{P(F)}$, so it is not the probability of some event in the space, but rather a ratio of two probabilities, which can't be represented on the Venn diagram.
– астон вілла олоф мэллбэрг
Nov 16 at 6:30
Is $P(E/F)$ a conditional probability. If so, then recall the definition of conditional probability.
– Lord Shark the Unknown
Nov 16 at 6:31
Is it not possible to sketch conditional probability on venn diagram @LordSharktheUnknown
– Abcd
Nov 16 at 6:33
@Abcd You can sketch it. Just cut out the circle for $F$. Now, this is your new sample space.
– trancelocation
Nov 16 at 7:17
You probably confuse the relative complement and the conditional probability.
– Hanul Jeon
Nov 16 at 6:30
You probably confuse the relative complement and the conditional probability.
– Hanul Jeon
Nov 16 at 6:30
You are confusing $/$ , that is conditional probability, with $cap$, the intersection. Conditional probability $P(E / F)$ requires that you take the ratio $frac{P(E cap F)}{P(F)}$, so it is not the probability of some event in the space, but rather a ratio of two probabilities, which can't be represented on the Venn diagram.
– астон вілла олоф мэллбэрг
Nov 16 at 6:30
You are confusing $/$ , that is conditional probability, with $cap$, the intersection. Conditional probability $P(E / F)$ requires that you take the ratio $frac{P(E cap F)}{P(F)}$, so it is not the probability of some event in the space, but rather a ratio of two probabilities, which can't be represented on the Venn diagram.
– астон вілла олоф мэллбэрг
Nov 16 at 6:30
Is $P(E/F)$ a conditional probability. If so, then recall the definition of conditional probability.
– Lord Shark the Unknown
Nov 16 at 6:31
Is $P(E/F)$ a conditional probability. If so, then recall the definition of conditional probability.
– Lord Shark the Unknown
Nov 16 at 6:31
Is it not possible to sketch conditional probability on venn diagram @LordSharktheUnknown
– Abcd
Nov 16 at 6:33
Is it not possible to sketch conditional probability on venn diagram @LordSharktheUnknown
– Abcd
Nov 16 at 6:33
@Abcd You can sketch it. Just cut out the circle for $F$. Now, this is your new sample space.
– trancelocation
Nov 16 at 7:17
@Abcd You can sketch it. Just cut out the circle for $F$. Now, this is your new sample space.
– trancelocation
Nov 16 at 7:17
add a comment |
2 Answers
2
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up vote
0
down vote
Note that
$(E cap bar F) cup (bar E cap bar F) = bar F$ and- $(E cap bar F) cap (bar E cap bar F) = emptyset$
answered Nov 16 at 6:37
trancelocation
8,4471520
add a comment |
up vote
0
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Since $E$ and $bar{E}$ are disjoint events we have:
$$mathbb{P}(E|bar{F}) + mathbb{P}(bar{E}|bar{F}) = mathbb{P}(E cup bar{E}|bar{F}) = mathbb{P}(Omega|bar{F}) = 1.$$
The first step follows from the additivity axiom of probability and the last follows from the norming axiom. Your diagram does not accurately show the conditional probabilities: the first is the relative size of the red area in the union of both the red and blue areas, and the second is the relative size of the blue area in the union of both the red and blue areas, which sum to the whole area (i.e., one).
answered Nov 16 at 6:54
Ben
88611
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Note that
$(E cap bar F) cup (bar E cap bar F) = bar F$ and- $(E cap bar F) cap (bar E cap bar F) = emptyset$
answered Nov 16 at 6:37
trancelocation
8,4471520
add a comment |
up vote
0
down vote
Note that
$(E cap bar F) cup (bar E cap bar F) = bar F$ and- $(E cap bar F) cap (bar E cap bar F) = emptyset$
answered Nov 16 at 6:37
trancelocation
8,4471520
add a comment |
up vote
0
down vote
up vote
0
down vote
Note that
$(E cap bar F) cup (bar E cap bar F) = bar F$ and- $(E cap bar F) cap (bar E cap bar F) = emptyset$
answered Nov 16 at 6:37
trancelocation
8,4471520
Note that
$(E cap bar F) cup (bar E cap bar F) = bar F$ and- $(E cap bar F) cap (bar E cap bar F) = emptyset$
answered Nov 16 at 6:37
trancelocation
8,4471520
answered Nov 16 at 6:37
trancelocation
8,4471520
answered Nov 16 at 6:37
trancelocation
8,4471520
answered Nov 16 at 6:37
trancelocation
8,4471520
8,4471520
add a comment |
add a comment |
up vote
0
down vote
Since $E$ and $bar{E}$ are disjoint events we have:
$$mathbb{P}(E|bar{F}) + mathbb{P}(bar{E}|bar{F}) = mathbb{P}(E cup bar{E}|bar{F}) = mathbb{P}(Omega|bar{F}) = 1.$$
The first step follows from the additivity axiom of probability and the last follows from the norming axiom. Your diagram does not accurately show the conditional probabilities: the first is the relative size of the red area in the union of both the red and blue areas, and the second is the relative size of the blue area in the union of both the red and blue areas, which sum to the whole area (i.e., one).
answered Nov 16 at 6:54
Ben
88611
add a comment |
up vote
0
down vote
Since $E$ and $bar{E}$ are disjoint events we have:
$$mathbb{P}(E|bar{F}) + mathbb{P}(bar{E}|bar{F}) = mathbb{P}(E cup bar{E}|bar{F}) = mathbb{P}(Omega|bar{F}) = 1.$$
The first step follows from the additivity axiom of probability and the last follows from the norming axiom. Your diagram does not accurately show the conditional probabilities: the first is the relative size of the red area in the union of both the red and blue areas, and the second is the relative size of the blue area in the union of both the red and blue areas, which sum to the whole area (i.e., one).
answered Nov 16 at 6:54
Ben
88611
add a comment |
up vote
0
down vote
up vote
0
down vote
Since $E$ and $bar{E}$ are disjoint events we have:
$$mathbb{P}(E|bar{F}) + mathbb{P}(bar{E}|bar{F}) = mathbb{P}(E cup bar{E}|bar{F}) = mathbb{P}(Omega|bar{F}) = 1.$$
The first step follows from the additivity axiom of probability and the last follows from the norming axiom. Your diagram does not accurately show the conditional probabilities: the first is the relative size of the red area in the union of both the red and blue areas, and the second is the relative size of the blue area in the union of both the red and blue areas, which sum to the whole area (i.e., one).
answered Nov 16 at 6:54
Ben
88611
Since $E$ and $bar{E}$ are disjoint events we have:
$$mathbb{P}(E|bar{F}) + mathbb{P}(bar{E}|bar{F}) = mathbb{P}(E cup bar{E}|bar{F}) = mathbb{P}(Omega|bar{F}) = 1.$$
The first step follows from the additivity axiom of probability and the last follows from the norming axiom. Your diagram does not accurately show the conditional probabilities: the first is the relative size of the red area in the union of both the red and blue areas, and the second is the relative size of the blue area in the union of both the red and blue areas, which sum to the whole area (i.e., one).
answered Nov 16 at 6:54
Ben
88611
answered Nov 16 at 6:54
Ben
88611
answered Nov 16 at 6:54
Ben
88611
answered Nov 16 at 6:54
Ben
88611
88611
add a comment |
add a comment |
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You probably confuse the relative complement and the conditional probability.
– Hanul Jeon
Nov 16 at 6:30
You are confusing $/$ , that is conditional probability, with $cap$, the intersection. Conditional probability $P(E / F)$ requires that you take the ratio $frac{P(E cap F)}{P(F)}$, so it is not the probability of some event in the space, but rather a ratio of two probabilities, which can't be represented on the Venn diagram.
– астон вілла олоф мэллбэрг
Nov 16 at 6:30
Is $P(E/F)$ a conditional probability. If so, then recall the definition of conditional probability.
– Lord Shark the Unknown
Nov 16 at 6:31
Is it not possible to sketch conditional probability on venn diagram @LordSharktheUnknown
– Abcd
Nov 16 at 6:33
@Abcd You can sketch it. Just cut out the circle for $F$. Now, this is your new sample space.
– trancelocation
Nov 16 at 7:17