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 Mar 24 at 18:22
Robert Howard
2,2933935
asked Mar 24 at 5:48
Gordon NgGordon Ng
292
$endgroup$
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 Mar 24 at 18:22
Robert Howard
2,2933935
asked Mar 24 at 5:48
Gordon NgGordon Ng
292
$endgroup$
6
$begingroup$
I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
$endgroup$
– Paras Khosla
Mar 24 at 5:49
3
$begingroup$
A common way to show that something doesn’t hold is to come up with a specific counterexample.
$endgroup$
– amd
Mar 24 at 5:51
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
Mar 24 at 10:01
$begingroup$
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
Mar 24 at 18:23
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 Mar 24 at 18:22
Robert Howard
2,2933935
asked Mar 24 at 5:48
Gordon NgGordon Ng
292
$endgroup$
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 Mar 24 at 18:22
Robert Howard
2,2933935
asked Mar 24 at 5:48
Gordon NgGordon Ng
292
edited Mar 24 at 18:22
Robert Howard
2,2933935
asked Mar 24 at 5:48
Gordon NgGordon Ng
292
edited Mar 24 at 18:22
Robert Howard
2,2933935
edited Mar 24 at 18:22
Robert Howard
2,2933935
edited Mar 24 at 18:22
Robert Howard
2,2933935
2,2933935
asked Mar 24 at 5:48
Gordon NgGordon Ng
292
asked Mar 24 at 5:48
Gordon NgGordon Ng
292
asked Mar 24 at 5:48
Gordon NgGordon Ng
292
292
6
$begingroup$
I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
$endgroup$
– Paras Khosla
Mar 24 at 5:49
3
$begingroup$
A common way to show that something doesn’t hold is to come up with a specific counterexample.
$endgroup$
– amd
Mar 24 at 5:51
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
Mar 24 at 10:01
$begingroup$
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
Mar 24 at 18:23
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
Mar 24 at 5:49
3
$begingroup$
A common way to show that something doesn’t hold is to come up with a specific counterexample.
$endgroup$
– amd
Mar 24 at 5:51
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
Mar 24 at 10:01
$begingroup$
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
Mar 24 at 18:23
6
6
$begingroup$
I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
$endgroup$
– Paras Khosla
Mar 24 at 5:49
$begingroup$
I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
$endgroup$
– Paras Khosla
Mar 24 at 5:49
3
3
$begingroup$
A common way to show that something doesn’t hold is to come up with a specific counterexample.
$endgroup$
– amd
Mar 24 at 5:51
$begingroup$
A common way to show that something doesn’t hold is to come up with a specific counterexample.
$endgroup$
– amd
Mar 24 at 5:51
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
Mar 24 at 10:01
$begingroup$
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
Mar 24 at 10:01
$begingroup$
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
Mar 24 at 18:23
$begingroup$
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
Mar 24 at 18:23
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
Take any event of probability $frac 1 2$ and take $B=A$.
answered Mar 24 at 5:54
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
$endgroup$
$begingroup$
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
Mar 24 at 18:29
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 Mar 24 at 11:28
user985366user985366
432310
$endgroup$
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 Mar 24 at 5:55
TojrahTojrah
3836
$endgroup$
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 Mar 24 at 11:29
user21820user21820
40k544160
$endgroup$
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 Mar 24 at 18:08
Robert Howard
2,2933935
answered Mar 24 at 17:57
Artemis FowlArtemis Fowl
1114
$endgroup$
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 Mar 24 at 16:28
BakuriuBakuriu
11115
$endgroup$
add a comment |
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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 Mar 24 at 5:54
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
$endgroup$
$begingroup$
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
Mar 24 at 18:29
add a comment |
$begingroup$
Take any event of probability $frac 1 2$ and take $B=A$.
answered Mar 24 at 5:54
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
$endgroup$
$begingroup$
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
Mar 24 at 18:29
add a comment |
$begingroup$
Take any event of probability $frac 1 2$ and take $B=A$.
answered Mar 24 at 5:54
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
$endgroup$
Take any event of probability $frac 1 2$ and take $B=A$.
answered Mar 24 at 5:54
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
answered Mar 24 at 5:54
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
answered Mar 24 at 5:54
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
answered Mar 24 at 5:54
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
71.9k53170
$begingroup$
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
Mar 24 at 18:29
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
Mar 24 at 18:29
$begingroup$
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
Mar 24 at 18:29
$begingroup$
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
Mar 24 at 18:29
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 Mar 24 at 11:28
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 Mar 24 at 11:28
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 Mar 24 at 11:28
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 Mar 24 at 11:28
user985366user985366
432310
answered Mar 24 at 11:28
user985366user985366
432310
answered Mar 24 at 11:28
user985366user985366
432310
answered Mar 24 at 11:28
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 Mar 24 at 5:55
TojrahTojrah
3836
$endgroup$
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 Mar 24 at 5:55
TojrahTojrah
3836
$endgroup$
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 Mar 24 at 5:55
TojrahTojrah
3836
$endgroup$
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 Mar 24 at 5:55
TojrahTojrah
3836
answered Mar 24 at 5:55
TojrahTojrah
3836
answered Mar 24 at 5:55
TojrahTojrah
3836
answered Mar 24 at 5:55
TojrahTojrah
3836
3836
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 Mar 24 at 11:29
user21820user21820
40k544160
$endgroup$
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 Mar 24 at 11:29
user21820user21820
40k544160
$endgroup$
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 Mar 24 at 11:29
user21820user21820
40k544160
$endgroup$
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 Mar 24 at 11:29
user21820user21820
40k544160
answered Mar 24 at 11:29
user21820user21820
40k544160
answered Mar 24 at 11:29
user21820user21820
40k544160
answered Mar 24 at 11:29
user21820user21820
40k544160
40k544160
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 Mar 24 at 18:08
Robert Howard
2,2933935
answered Mar 24 at 17:57
Artemis FowlArtemis Fowl
1114
$endgroup$
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 Mar 24 at 18:08
Robert Howard
2,2933935
answered Mar 24 at 17:57
Artemis FowlArtemis Fowl
1114
$endgroup$
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 Mar 24 at 18:08
Robert Howard
2,2933935
answered Mar 24 at 17:57
Artemis FowlArtemis Fowl
1114
$endgroup$
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 Mar 24 at 18:08
Robert Howard
2,2933935
answered Mar 24 at 17:57
Artemis FowlArtemis Fowl
1114
edited Mar 24 at 18:08
Robert Howard
2,2933935
edited Mar 24 at 18:08
Robert Howard
2,2933935
edited Mar 24 at 18:08
Robert Howard
2,2933935
2,2933935
answered Mar 24 at 17:57
Artemis FowlArtemis Fowl
1114
answered Mar 24 at 17:57
Artemis FowlArtemis Fowl
1114
answered Mar 24 at 17:57
Artemis FowlArtemis Fowl
1114
1114
add a comment |
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 Mar 24 at 16:28
BakuriuBakuriu
11115
$endgroup$
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 Mar 24 at 16:28
BakuriuBakuriu
11115
$endgroup$
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 Mar 24 at 16:28
BakuriuBakuriu
11115
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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 Mar 24 at 16:28
BakuriuBakuriu
11115
answered Mar 24 at 16:28
BakuriuBakuriu
11115
answered Mar 24 at 16:28
BakuriuBakuriu
11115
answered Mar 24 at 16:28
BakuriuBakuriu
11115
11115
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6
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I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
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– Paras Khosla
Mar 24 at 5:49
3
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A common way to show that something doesn’t hold is to come up with a specific counterexample.
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– amd
Mar 24 at 5:51
7
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If you flip a coin and get tails, does it mean that the next person you'll meet cannot be a girl?
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– Eric Duminil
Mar 24 at 10:01
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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/…
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– Robert Howard
Mar 24 at 18:23