By means of an example, show that $P(A) + P(B) = 1$ does not mean that $B$ is the complement of $A$
$begingroup$
I'm in grade 10, and I've just started to learn about complementary events. I am rather perplexed with this question. Isn't this question kinda contradictory, since $P(A) + P(A') = 1$?
This is what I got to:
$P(A) + P(B) = 1$
$P(A) + P(A') = 1$
How could it be proven that $B$ isn't the complement of $A$?
Help would be greatly appreciated.
probability
edited yesterday
Robert Howard
2,2383935
asked yesterday
Gordon NgGordon Ng
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Gordon Ng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
I'm in grade 10, and I've just started to learn about complementary events. I am rather perplexed with this question. Isn't this question kinda contradictory, since $P(A) + P(A') = 1$?
This is what I got to:
$P(A) + P(B) = 1$
$P(A) + P(A') = 1$
How could it be proven that $B$ isn't the complement of $A$?
Help would be greatly appreciated.
probability
edited yesterday
Robert Howard
2,2383935
asked yesterday
Gordon NgGordon Ng
292
New contributor
Gordon Ng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
6
$begingroup$
I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
$endgroup$
– Paras Khosla
yesterday
3
$begingroup$
A common way to show that something doesn’t hold is to come up with a specific counterexample.
$endgroup$
– amd
yesterday
7
$begingroup$
If you flip a coin and get tails, does it mean that the next person you'll meet cannot be a girl?
$endgroup$
– Eric Duminil
yesterday
$begingroup$
Welcome to MSE; this is a good first question! Here's a brief tutorial on how we format math on this site, like I did in my edit to your question: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Robert Howard
yesterday
add a comment |
$begingroup$
I'm in grade 10, and I've just started to learn about complementary events. I am rather perplexed with this question. Isn't this question kinda contradictory, since $P(A) + P(A') = 1$?
This is what I got to:
$P(A) + P(B) = 1$
$P(A) + P(A') = 1$
How could it be proven that $B$ isn't the complement of $A$?
Help would be greatly appreciated.
probability
edited yesterday
Robert Howard
2,2383935
asked yesterday
Gordon NgGordon Ng
292
New contributor
Gordon Ng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm in grade 10, and I've just started to learn about complementary events. I am rather perplexed with this question. Isn't this question kinda contradictory, since $P(A) + P(A') = 1$?
This is what I got to:
$P(A) + P(B) = 1$
$P(A) + P(A') = 1$
How could it be proven that $B$ isn't the complement of $A$?
Help would be greatly appreciated.
probability
probability
edited yesterday
Robert Howard
2,2383935
asked yesterday
Gordon NgGordon Ng
292
New contributor
Gordon Ng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Robert Howard
2,2383935
asked yesterday
Gordon NgGordon Ng
292
New contributor
Gordon Ng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited yesterday
Robert Howard
2,2383935
edited yesterday
Robert Howard
2,2383935
edited yesterday
Robert Howard
2,2383935
2,2383935
asked yesterday
Gordon NgGordon Ng
292
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Gordon Ng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked yesterday
Gordon NgGordon Ng
292
asked yesterday
Gordon NgGordon Ng
292
292
New contributor
Gordon Ng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Gordon Ng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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6
$begingroup$
I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
$endgroup$
– Paras Khosla
yesterday
3
$begingroup$
A common way to show that something doesn’t hold is to come up with a specific counterexample.
$endgroup$
– amd
yesterday
7
$begingroup$
If you flip a coin and get tails, does it mean that the next person you'll meet cannot be a girl?
$endgroup$
– Eric Duminil
yesterday
$begingroup$
Welcome to MSE; this is a good first question! Here's a brief tutorial on how we format math on this site, like I did in my edit to your question: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Robert Howard
yesterday
add a comment |
6
$begingroup$
I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
$endgroup$
– Paras Khosla
yesterday
3
$begingroup$
A common way to show that something doesn’t hold is to come up with a specific counterexample.
$endgroup$
– amd
yesterday
7
$begingroup$
If you flip a coin and get tails, does it mean that the next person you'll meet cannot be a girl?
$endgroup$
– Eric Duminil
yesterday
$begingroup$
Welcome to MSE; this is a good first question! Here's a brief tutorial on how we format math on this site, like I did in my edit to your question: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Robert Howard
yesterday
6
6
$begingroup$
I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
$endgroup$
– Paras Khosla
yesterday
$begingroup$
I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
$endgroup$
– Paras Khosla
yesterday
3
3
$begingroup$
A common way to show that something doesn’t hold is to come up with a specific counterexample.
$endgroup$
– amd
yesterday
$begingroup$
A common way to show that something doesn’t hold is to come up with a specific counterexample.
$endgroup$
– amd
yesterday
7
7
$begingroup$
If you flip a coin and get tails, does it mean that the next person you'll meet cannot be a girl?
$endgroup$
– Eric Duminil
yesterday
$begingroup$
If you flip a coin and get tails, does it mean that the next person you'll meet cannot be a girl?
$endgroup$
– Eric Duminil
yesterday
$begingroup$
Welcome to MSE; this is a good first question! Here's a brief tutorial on how we format math on this site, like I did in my edit to your question: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Robert Howard
yesterday
$begingroup$
Welcome to MSE; this is a good first question! Here's a brief tutorial on how we format math on this site, like I did in my edit to your question: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Robert Howard
yesterday
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
Take any event of probability $frac 1 2$ and take $B=A$.
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
$endgroup$
$begingroup$
And if it helps to give a more concrete example, we can say: I'm going to flip a coin just once. Let $A$ be the event that the flip will be heads, and let $B$ also be the event that the flip will be heads.
$endgroup$
– aschepler
yesterday
add a comment |
$begingroup$
A counterexample
Take a normal 6-sided die.
Let event A be "roll any of the numbers 1, 2, 3 or 4". P(A) = 4/6
Let event B be "roll any of the numbers 1 or 2". P(B) = 2/6
P(A) + P(B) = 4/6 + 2/6 = 6/6 = 1
But B is not the complement of A.
The complement of A is the event "roll any of the numbers 5 or 6".
By this example, we've shown that P(A) + P(B) = 1 does not imply that A and B are complements.
Further
It also seems you have misunderstood the question.
You wrote How could it be proven that B isn't the complement of A?
This is not what you need to prove, and you cannot prove it just by knowing the probabilities. What you need to show is that it isn't always the case. You can show this by giving 1 counterexample, as above.
answered yesterday
user985366user985366
432310
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add a comment |
$begingroup$
As commented by @amd you can have many counter examples such as in throwing of dice. You can define any two events A and B. Such that n(A) + n(B)=6. So P(A)+P(B)=1. But A and B need not be necessarily disjoint sets.
answered yesterday
TojrahTojrah
2455
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From the two statements you obtained (correctly), you can further obtain $P(A') = P(B)$. But that does not imply $A'=B$. Just as $x^2 = y^2$ for reals $x,y$ does not imply $x = y$.
answered yesterday
user21820user21820
39.7k544158
$endgroup$
add a comment |
$begingroup$
Another example:
$$A = text{Getting a head on coin }A$$
$$B = text{Getting a head on coin }B$$
Both have probability $0.5$, so $P(A)+P(B)=1$, but you could get a head on either, both or neither.
edited yesterday
Robert Howard
2,2383935
answered yesterday
Shadow The GohstShadow The Gohst
111
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Let $S$ be any infinite set and let $e in S$. Now let $A$ and $B$ be a partition of $S setminus {e}$ with both $A$ and $B$ infinite. Let $P(C)$ be the probability of drawing randomly an element of $S$ that is in the set $C$.
Since $S = A cup B cup {e}$ we have:
$$
P(A) + P(B) + P({e}) = 1
$$
but also
$$
P({e}) = 0
$$
since it is finite, thus:
$$
P(A) + P(B) = 1
$$
For example $S = mathbb{N}$, $e={0}$, $A={x mid x in mathbb{N}_0 wedge x text{ is even}}$ and $B={x mid x in mathbb{N}_0 wedge x text{ is odd}}$.
answered yesterday
BakuriuBakuriu
11115
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take any event of probability $frac 1 2$ and take $B=A$.
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
$endgroup$
$begingroup$
And if it helps to give a more concrete example, we can say: I'm going to flip a coin just once. Let $A$ be the event that the flip will be heads, and let $B$ also be the event that the flip will be heads.
$endgroup$
– aschepler
yesterday
add a comment |
$begingroup$
Take any event of probability $frac 1 2$ and take $B=A$.
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
$endgroup$
$begingroup$
And if it helps to give a more concrete example, we can say: I'm going to flip a coin just once. Let $A$ be the event that the flip will be heads, and let $B$ also be the event that the flip will be heads.
$endgroup$
– aschepler
yesterday
add a comment |
$begingroup$
Take any event of probability $frac 1 2$ and take $B=A$.
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
$endgroup$
Take any event of probability $frac 1 2$ and take $B=A$.
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
69.9k53170
$begingroup$
And if it helps to give a more concrete example, we can say: I'm going to flip a coin just once. Let $A$ be the event that the flip will be heads, and let $B$ also be the event that the flip will be heads.
$endgroup$
– aschepler
yesterday
add a comment |
$begingroup$
And if it helps to give a more concrete example, we can say: I'm going to flip a coin just once. Let $A$ be the event that the flip will be heads, and let $B$ also be the event that the flip will be heads.
$endgroup$
– aschepler
yesterday
$begingroup$
And if it helps to give a more concrete example, we can say: I'm going to flip a coin just once. Let $A$ be the event that the flip will be heads, and let $B$ also be the event that the flip will be heads.
$endgroup$
– aschepler
yesterday
$begingroup$
And if it helps to give a more concrete example, we can say: I'm going to flip a coin just once. Let $A$ be the event that the flip will be heads, and let $B$ also be the event that the flip will be heads.
$endgroup$
– aschepler
yesterday
add a comment |
$begingroup$
A counterexample
Take a normal 6-sided die.
Let event A be "roll any of the numbers 1, 2, 3 or 4". P(A) = 4/6
Let event B be "roll any of the numbers 1 or 2". P(B) = 2/6
P(A) + P(B) = 4/6 + 2/6 = 6/6 = 1
But B is not the complement of A.
The complement of A is the event "roll any of the numbers 5 or 6".
By this example, we've shown that P(A) + P(B) = 1 does not imply that A and B are complements.
Further
It also seems you have misunderstood the question.
You wrote How could it be proven that B isn't the complement of A?
This is not what you need to prove, and you cannot prove it just by knowing the probabilities. What you need to show is that it isn't always the case. You can show this by giving 1 counterexample, as above.
answered yesterday
user985366user985366
432310
$endgroup$
add a comment |
$begingroup$
A counterexample
Take a normal 6-sided die.
Let event A be "roll any of the numbers 1, 2, 3 or 4". P(A) = 4/6
Let event B be "roll any of the numbers 1 or 2". P(B) = 2/6
P(A) + P(B) = 4/6 + 2/6 = 6/6 = 1
But B is not the complement of A.
The complement of A is the event "roll any of the numbers 5 or 6".
By this example, we've shown that P(A) + P(B) = 1 does not imply that A and B are complements.
Further
It also seems you have misunderstood the question.
You wrote How could it be proven that B isn't the complement of A?
This is not what you need to prove, and you cannot prove it just by knowing the probabilities. What you need to show is that it isn't always the case. You can show this by giving 1 counterexample, as above.
answered yesterday
user985366user985366
432310
$endgroup$
add a comment |
$begingroup$
A counterexample
Take a normal 6-sided die.
Let event A be "roll any of the numbers 1, 2, 3 or 4". P(A) = 4/6
Let event B be "roll any of the numbers 1 or 2". P(B) = 2/6
P(A) + P(B) = 4/6 + 2/6 = 6/6 = 1
But B is not the complement of A.
The complement of A is the event "roll any of the numbers 5 or 6".
By this example, we've shown that P(A) + P(B) = 1 does not imply that A and B are complements.
Further
It also seems you have misunderstood the question.
You wrote How could it be proven that B isn't the complement of A?
This is not what you need to prove, and you cannot prove it just by knowing the probabilities. What you need to show is that it isn't always the case. You can show this by giving 1 counterexample, as above.
answered yesterday
user985366user985366
432310
$endgroup$
A counterexample
Take a normal 6-sided die.
Let event A be "roll any of the numbers 1, 2, 3 or 4". P(A) = 4/6
Let event B be "roll any of the numbers 1 or 2". P(B) = 2/6
P(A) + P(B) = 4/6 + 2/6 = 6/6 = 1
But B is not the complement of A.
The complement of A is the event "roll any of the numbers 5 or 6".
By this example, we've shown that P(A) + P(B) = 1 does not imply that A and B are complements.
Further
It also seems you have misunderstood the question.
You wrote How could it be proven that B isn't the complement of A?
This is not what you need to prove, and you cannot prove it just by knowing the probabilities. What you need to show is that it isn't always the case. You can show this by giving 1 counterexample, as above.
answered yesterday
user985366user985366
432310
answered yesterday
user985366user985366
432310
answered yesterday
user985366user985366
432310
answered yesterday
user985366user985366
432310
432310
add a comment |
add a comment |
$begingroup$
As commented by @amd you can have many counter examples such as in throwing of dice. You can define any two events A and B. Such that n(A) + n(B)=6. So P(A)+P(B)=1. But A and B need not be necessarily disjoint sets.
answered yesterday
TojrahTojrah
2455
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add a comment |
$begingroup$
As commented by @amd you can have many counter examples such as in throwing of dice. You can define any two events A and B. Such that n(A) + n(B)=6. So P(A)+P(B)=1. But A and B need not be necessarily disjoint sets.
answered yesterday
TojrahTojrah
2455
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Tojrah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
As commented by @amd you can have many counter examples such as in throwing of dice. You can define any two events A and B. Such that n(A) + n(B)=6. So P(A)+P(B)=1. But A and B need not be necessarily disjoint sets.
answered yesterday
TojrahTojrah
2455
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Tojrah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
As commented by @amd you can have many counter examples such as in throwing of dice. You can define any two events A and B. Such that n(A) + n(B)=6. So P(A)+P(B)=1. But A and B need not be necessarily disjoint sets.
answered yesterday
TojrahTojrah
2455
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Tojrah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered yesterday
TojrahTojrah
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answered yesterday
TojrahTojrah
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answered yesterday
TojrahTojrah
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add a comment |
add a comment |
$begingroup$
From the two statements you obtained (correctly), you can further obtain $P(A') = P(B)$. But that does not imply $A'=B$. Just as $x^2 = y^2$ for reals $x,y$ does not imply $x = y$.
answered yesterday
user21820user21820
39.7k544158
$endgroup$
add a comment |
$begingroup$
From the two statements you obtained (correctly), you can further obtain $P(A') = P(B)$. But that does not imply $A'=B$. Just as $x^2 = y^2$ for reals $x,y$ does not imply $x = y$.
answered yesterday
user21820user21820
39.7k544158
$endgroup$
add a comment |
$begingroup$
From the two statements you obtained (correctly), you can further obtain $P(A') = P(B)$. But that does not imply $A'=B$. Just as $x^2 = y^2$ for reals $x,y$ does not imply $x = y$.
answered yesterday
user21820user21820
39.7k544158
$endgroup$
From the two statements you obtained (correctly), you can further obtain $P(A') = P(B)$. But that does not imply $A'=B$. Just as $x^2 = y^2$ for reals $x,y$ does not imply $x = y$.
answered yesterday
user21820user21820
39.7k544158
answered yesterday
user21820user21820
39.7k544158
answered yesterday
user21820user21820
39.7k544158
answered yesterday
user21820user21820
39.7k544158
39.7k544158
add a comment |
add a comment |
$begingroup$
Another example:
$$A = text{Getting a head on coin }A$$
$$B = text{Getting a head on coin }B$$
Both have probability $0.5$, so $P(A)+P(B)=1$, but you could get a head on either, both or neither.
edited yesterday
Robert Howard
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answered yesterday
Shadow The GohstShadow The Gohst
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Another example:
$$A = text{Getting a head on coin }A$$
$$B = text{Getting a head on coin }B$$
Both have probability $0.5$, so $P(A)+P(B)=1$, but you could get a head on either, both or neither.
edited yesterday
Robert Howard
2,2383935
answered yesterday
Shadow The GohstShadow The Gohst
111
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Shadow The Gohst is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$begingroup$
Another example:
$$A = text{Getting a head on coin }A$$
$$B = text{Getting a head on coin }B$$
Both have probability $0.5$, so $P(A)+P(B)=1$, but you could get a head on either, both or neither.
edited yesterday
Robert Howard
2,2383935
answered yesterday
Shadow The GohstShadow The Gohst
111
New contributor
Shadow The Gohst is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
Another example:
$$A = text{Getting a head on coin }A$$
$$B = text{Getting a head on coin }B$$
Both have probability $0.5$, so $P(A)+P(B)=1$, but you could get a head on either, both or neither.
edited yesterday
Robert Howard
2,2383935
answered yesterday
Shadow The GohstShadow The Gohst
111
New contributor
Shadow The Gohst is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Robert Howard
2,2383935
edited yesterday
Robert Howard
2,2383935
edited yesterday
Robert Howard
2,2383935
2,2383935
answered yesterday
Shadow The GohstShadow The Gohst
111
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$begingroup$
Let $S$ be any infinite set and let $e in S$. Now let $A$ and $B$ be a partition of $S setminus {e}$ with both $A$ and $B$ infinite. Let $P(C)$ be the probability of drawing randomly an element of $S$ that is in the set $C$.
Since $S = A cup B cup {e}$ we have:
$$
P(A) + P(B) + P({e}) = 1
$$
but also
$$
P({e}) = 0
$$
since it is finite, thus:
$$
P(A) + P(B) = 1
$$
For example $S = mathbb{N}$, $e={0}$, $A={x mid x in mathbb{N}_0 wedge x text{ is even}}$ and $B={x mid x in mathbb{N}_0 wedge x text{ is odd}}$.
answered yesterday
BakuriuBakuriu
11115
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$endgroup$
add a comment |
$begingroup$
Let $S$ be any infinite set and let $e in S$. Now let $A$ and $B$ be a partition of $S setminus {e}$ with both $A$ and $B$ infinite. Let $P(C)$ be the probability of drawing randomly an element of $S$ that is in the set $C$.
Since $S = A cup B cup {e}$ we have:
$$
P(A) + P(B) + P({e}) = 1
$$
but also
$$
P({e}) = 0
$$
since it is finite, thus:
$$
P(A) + P(B) = 1
$$
For example $S = mathbb{N}$, $e={0}$, $A={x mid x in mathbb{N}_0 wedge x text{ is even}}$ and $B={x mid x in mathbb{N}_0 wedge x text{ is odd}}$.
answered yesterday
BakuriuBakuriu
11115
New contributor
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Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Let $S$ be any infinite set and let $e in S$. Now let $A$ and $B$ be a partition of $S setminus {e}$ with both $A$ and $B$ infinite. Let $P(C)$ be the probability of drawing randomly an element of $S$ that is in the set $C$.
Since $S = A cup B cup {e}$ we have:
$$
P(A) + P(B) + P({e}) = 1
$$
but also
$$
P({e}) = 0
$$
since it is finite, thus:
$$
P(A) + P(B) = 1
$$
For example $S = mathbb{N}$, $e={0}$, $A={x mid x in mathbb{N}_0 wedge x text{ is even}}$ and $B={x mid x in mathbb{N}_0 wedge x text{ is odd}}$.
answered yesterday
BakuriuBakuriu
11115
New contributor
Bakuriu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let $S$ be any infinite set and let $e in S$. Now let $A$ and $B$ be a partition of $S setminus {e}$ with both $A$ and $B$ infinite. Let $P(C)$ be the probability of drawing randomly an element of $S$ that is in the set $C$.
Since $S = A cup B cup {e}$ we have:
$$
P(A) + P(B) + P({e}) = 1
$$
but also
$$
P({e}) = 0
$$
since it is finite, thus:
$$
P(A) + P(B) = 1
$$
For example $S = mathbb{N}$, $e={0}$, $A={x mid x in mathbb{N}_0 wedge x text{ is even}}$ and $B={x mid x in mathbb{N}_0 wedge x text{ is odd}}$.
answered yesterday
BakuriuBakuriu
11115
New contributor
Bakuriu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
BakuriuBakuriu
11115
New contributor
Bakuriu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered yesterday
BakuriuBakuriu
11115
answered yesterday
BakuriuBakuriu
11115
11115
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Gordon Ng is a new contributor. Be nice, and check out our Code of Conduct.
Gordon Ng is a new contributor. Be nice, and check out our Code of Conduct.
Gordon Ng is a new contributor. Be nice, and check out our Code of Conduct.
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6
$begingroup$
I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
$endgroup$
– Paras Khosla
yesterday
3
$begingroup$
A common way to show that something doesn’t hold is to come up with a specific counterexample.
$endgroup$
– amd
yesterday
7
$begingroup$
If you flip a coin and get tails, does it mean that the next person you'll meet cannot be a girl?
$endgroup$
– Eric Duminil
yesterday
$begingroup$
Welcome to MSE; this is a good first question! Here's a brief tutorial on how we format math on this site, like I did in my edit to your question: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Robert Howard
yesterday