Let $A$ and $B$ be events. Is it necessarily true that $frac{P(A cap B)}{P(A)} = frac{P(A cap B)}{P(B)}$?
$begingroup$
Let $A$ and $B$ be events. Is it necessarily true that $dfrac{P(A cap B)}{P(A)} = dfrac{P(A cap B)}{P(B)}$?
If so, explain why. If not, give a counterexample.
I'm having a hard time proving this problem. Where should I start? How should I explain it in a way that makes sense and is a valid proof at the same time? Listing random events wouldn't help since it must be true for all events. Can someone walk me through this? Thanks!
probability combinatorics conditional-probability
edited Jan 1 at 16:12
Martin Sleziak
44.8k10119272
asked Oct 3 '16 at 22:26
Yuna KunYuna Kun
631218
$endgroup$
add a comment |
$begingroup$
Let $A$ and $B$ be events. Is it necessarily true that $dfrac{P(A cap B)}{P(A)} = dfrac{P(A cap B)}{P(B)}$?
If so, explain why. If not, give a counterexample.
I'm having a hard time proving this problem. Where should I start? How should I explain it in a way that makes sense and is a valid proof at the same time? Listing random events wouldn't help since it must be true for all events. Can someone walk me through this? Thanks!
probability combinatorics conditional-probability
edited Jan 1 at 16:12
Martin Sleziak
44.8k10119272
asked Oct 3 '16 at 22:26
Yuna KunYuna Kun
631218
$endgroup$
3
$begingroup$
Did you mean to switch $B$ and $A$ in the second fraction? If not, just divide both sides by $P(Acap B)$ and find some $P(A)ne P(B)$
$endgroup$
– Simply Beautiful Art
Oct 3 '16 at 22:28
$begingroup$
@SimpleArt Sorry, is the equation I wrote equivalent to $P(A|B) = P(B|A)$?
$endgroup$
– Yuna Kun
Oct 3 '16 at 22:32
$begingroup$
what if P(A) = 1 and P(B) = 1/2 and we assume independence - then you get $P(A cap B) = 1/2$ so LHS = 1/2 and RHS = 1
$endgroup$
– Cato
Oct 4 '16 at 9:38
$begingroup$
A way of explaining it to be not true is that for independent events the equation boils down to P(A) = P(B)
$endgroup$
– Cato
Oct 4 '16 at 9:43
add a comment |
$begingroup$
Let $A$ and $B$ be events. Is it necessarily true that $dfrac{P(A cap B)}{P(A)} = dfrac{P(A cap B)}{P(B)}$?
If so, explain why. If not, give a counterexample.
I'm having a hard time proving this problem. Where should I start? How should I explain it in a way that makes sense and is a valid proof at the same time? Listing random events wouldn't help since it must be true for all events. Can someone walk me through this? Thanks!
probability combinatorics conditional-probability
edited Jan 1 at 16:12
Martin Sleziak
44.8k10119272
asked Oct 3 '16 at 22:26
Yuna KunYuna Kun
631218
$endgroup$
Let $A$ and $B$ be events. Is it necessarily true that $dfrac{P(A cap B)}{P(A)} = dfrac{P(A cap B)}{P(B)}$?
If so, explain why. If not, give a counterexample.
I'm having a hard time proving this problem. Where should I start? How should I explain it in a way that makes sense and is a valid proof at the same time? Listing random events wouldn't help since it must be true for all events. Can someone walk me through this? Thanks!
probability combinatorics conditional-probability
probability combinatorics conditional-probability
edited Jan 1 at 16:12
Martin Sleziak
44.8k10119272
asked Oct 3 '16 at 22:26
Yuna KunYuna Kun
631218
edited Jan 1 at 16:12
Martin Sleziak
44.8k10119272
asked Oct 3 '16 at 22:26
Yuna KunYuna Kun
631218
edited Jan 1 at 16:12
Martin Sleziak
44.8k10119272
edited Jan 1 at 16:12
Martin Sleziak
44.8k10119272
edited Jan 1 at 16:12
Martin Sleziak
44.8k10119272
44.8k10119272
asked Oct 3 '16 at 22:26
Yuna KunYuna Kun
631218
asked Oct 3 '16 at 22:26
Yuna KunYuna Kun
631218
asked Oct 3 '16 at 22:26
Yuna KunYuna Kun
631218
631218
3
$begingroup$
Did you mean to switch $B$ and $A$ in the second fraction? If not, just divide both sides by $P(Acap B)$ and find some $P(A)ne P(B)$
$endgroup$
– Simply Beautiful Art
Oct 3 '16 at 22:28
$begingroup$
@SimpleArt Sorry, is the equation I wrote equivalent to $P(A|B) = P(B|A)$?
$endgroup$
– Yuna Kun
Oct 3 '16 at 22:32
$begingroup$
what if P(A) = 1 and P(B) = 1/2 and we assume independence - then you get $P(A cap B) = 1/2$ so LHS = 1/2 and RHS = 1
$endgroup$
– Cato
Oct 4 '16 at 9:38
$begingroup$
A way of explaining it to be not true is that for independent events the equation boils down to P(A) = P(B)
$endgroup$
– Cato
Oct 4 '16 at 9:43
add a comment |
3
$begingroup$
Did you mean to switch $B$ and $A$ in the second fraction? If not, just divide both sides by $P(Acap B)$ and find some $P(A)ne P(B)$
$endgroup$
– Simply Beautiful Art
Oct 3 '16 at 22:28
$begingroup$
@SimpleArt Sorry, is the equation I wrote equivalent to $P(A|B) = P(B|A)$?
$endgroup$
– Yuna Kun
Oct 3 '16 at 22:32
$begingroup$
what if P(A) = 1 and P(B) = 1/2 and we assume independence - then you get $P(A cap B) = 1/2$ so LHS = 1/2 and RHS = 1
$endgroup$
– Cato
Oct 4 '16 at 9:38
$begingroup$
A way of explaining it to be not true is that for independent events the equation boils down to P(A) = P(B)
$endgroup$
– Cato
Oct 4 '16 at 9:43
3
3
$begingroup$
Did you mean to switch $B$ and $A$ in the second fraction? If not, just divide both sides by $P(Acap B)$ and find some $P(A)ne P(B)$
$endgroup$
– Simply Beautiful Art
Oct 3 '16 at 22:28
$begingroup$
Did you mean to switch $B$ and $A$ in the second fraction? If not, just divide both sides by $P(Acap B)$ and find some $P(A)ne P(B)$
$endgroup$
– Simply Beautiful Art
Oct 3 '16 at 22:28
$begingroup$
@SimpleArt Sorry, is the equation I wrote equivalent to $P(A|B) = P(B|A)$?
$endgroup$
– Yuna Kun
Oct 3 '16 at 22:32
$begingroup$
@SimpleArt Sorry, is the equation I wrote equivalent to $P(A|B) = P(B|A)$?
$endgroup$
– Yuna Kun
Oct 3 '16 at 22:32
$begingroup$
what if P(A) = 1 and P(B) = 1/2 and we assume independence - then you get $P(A cap B) = 1/2$ so LHS = 1/2 and RHS = 1
$endgroup$
– Cato
Oct 4 '16 at 9:38
$begingroup$
what if P(A) = 1 and P(B) = 1/2 and we assume independence - then you get $P(A cap B) = 1/2$ so LHS = 1/2 and RHS = 1
$endgroup$
– Cato
Oct 4 '16 at 9:38
$begingroup$
A way of explaining it to be not true is that for independent events the equation boils down to P(A) = P(B)
$endgroup$
– Cato
Oct 4 '16 at 9:43
$begingroup$
A way of explaining it to be not true is that for independent events the equation boils down to P(A) = P(B)
$endgroup$
– Cato
Oct 4 '16 at 9:43
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Hint:
$P(A|B)$ need not be equal to $P(B|A)$.
Let $A$ be the event that you get number $2$ from a dice toss.
Let $B$ be the event that you get an even number from the same dice toss.
answered Oct 3 '16 at 22:33
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
add a comment |
$begingroup$
Say you have a standard six-sided die and a fair coin. You roll first the die and get $1,2,3,4,5,6$ with equal prob. and then throw the coin and get $H,T$ with equal prob.
Define $A$ to be "get a $1$ with the die" and $B$ "get a $Head$ with the coin".
Clearly $P(A) = 1/6$ and $P(B) = 1/2$ and both event are independent. Therefore
$$
frac{P(A cap B)}{P(A)} = frac{1/12}{1/6} neq frac{1/12}{1/2} = frac{P(A cap B)}{P(B)}
$$
answered Oct 3 '16 at 22:32
ZubzubZubzub
3,8891922
$endgroup$
add a comment |
$begingroup$
Since
$$P(Acap B)=P(A)P(B|A)=P(B)P(A|B)$$
We can see
$$frac{P(Acap B)}{P(A)}=frac{P(A)P(B|A)}{P(A)}=P(B|A)$$
but
$$frac{P(Acap B)}{P(B)}=frac{P(B)P(A|B)}{P(B)}=P(A|B)$$
Hence,
$$frac{P(Acap B)}{P(B)}neqfrac{P(Acap B)}{P(A)}$$
The example is when $A$ and $B$ are independent. In such case $P(A|B)=P(A)$ but $P(B|A)=P(B)$
answered Oct 3 '16 at 22:33
msmmsm
6,2292829
$endgroup$
add a comment |
$begingroup$
Honestly, i think it does not(i might be wrong, but here we go).
Notice that $frac{P(A cap B)}{P(A)} $ is actually the probability of the event $B$ knowing that the event $A$ happened. (Here, your sample space is actually the event $A$, so this formula is nothing but $P(C) = frac{C}{S}$ where $C$ is an event and $S$ the sample space.
Now, think about it:
Is the probability of an event $A$, knowing that $B$ has happened, equal to the probability of an event $B$, knowing that $A$ happened? The answer is no when $P(A)$ is not equal to $P(B)$. And that can be easily demonstrated:
You have:
$$ frac{P(A cap B)}{P(A)} = frac{P(A cap B)}{P(B)} $$
Obviously, this equality only holds if
$$P(A) = P(B) $$.
An example:
I have two dices, one black an one white. Let A be the event of the black resulting in 6 and B the event of the white resulting in 6.
I throw the first dice, so i have:
$$P(A) = 1/6 $$
And then:
$$P(B) = 1/6 $$
Since both of the events are independent, we have that
$$P(A cap B) = P(A)P(B) $$
So:
$$P(A/B) = P(A) = P(B/A) = P(B) $$.
This is an example that the equality holds for $P(A) = P(B)$. Any case that this doesn't apply, the equality doesn't apply.
answered Oct 3 '16 at 22:42
Vitor C GoergenVitor C Goergen
8261416
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
$P(A|B)$ need not be equal to $P(B|A)$.
Let $A$ be the event that you get number $2$ from a dice toss.
Let $B$ be the event that you get an even number from the same dice toss.
answered Oct 3 '16 at 22:33
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
add a comment |
$begingroup$
Hint:
$P(A|B)$ need not be equal to $P(B|A)$.
Let $A$ be the event that you get number $2$ from a dice toss.
Let $B$ be the event that you get an even number from the same dice toss.
answered Oct 3 '16 at 22:33
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
add a comment |
$begingroup$
Hint:
$P(A|B)$ need not be equal to $P(B|A)$.
Let $A$ be the event that you get number $2$ from a dice toss.
Let $B$ be the event that you get an even number from the same dice toss.
answered Oct 3 '16 at 22:33
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
Hint:
$P(A|B)$ need not be equal to $P(B|A)$.
Let $A$ be the event that you get number $2$ from a dice toss.
Let $B$ be the event that you get an even number from the same dice toss.
answered Oct 3 '16 at 22:33
Siong Thye GohSiong Thye Goh
102k1466118
answered Oct 3 '16 at 22:33
Siong Thye GohSiong Thye Goh
102k1466118
answered Oct 3 '16 at 22:33
Siong Thye GohSiong Thye Goh
102k1466118
answered Oct 3 '16 at 22:33
Siong Thye GohSiong Thye Goh
102k1466118
102k1466118
add a comment |
add a comment |
$begingroup$
Say you have a standard six-sided die and a fair coin. You roll first the die and get $1,2,3,4,5,6$ with equal prob. and then throw the coin and get $H,T$ with equal prob.
Define $A$ to be "get a $1$ with the die" and $B$ "get a $Head$ with the coin".
Clearly $P(A) = 1/6$ and $P(B) = 1/2$ and both event are independent. Therefore
$$
frac{P(A cap B)}{P(A)} = frac{1/12}{1/6} neq frac{1/12}{1/2} = frac{P(A cap B)}{P(B)}
$$
answered Oct 3 '16 at 22:32
ZubzubZubzub
3,8891922
$endgroup$
add a comment |
$begingroup$
Say you have a standard six-sided die and a fair coin. You roll first the die and get $1,2,3,4,5,6$ with equal prob. and then throw the coin and get $H,T$ with equal prob.
Define $A$ to be "get a $1$ with the die" and $B$ "get a $Head$ with the coin".
Clearly $P(A) = 1/6$ and $P(B) = 1/2$ and both event are independent. Therefore
$$
frac{P(A cap B)}{P(A)} = frac{1/12}{1/6} neq frac{1/12}{1/2} = frac{P(A cap B)}{P(B)}
$$
answered Oct 3 '16 at 22:32
ZubzubZubzub
3,8891922
$endgroup$
add a comment |
$begingroup$
Say you have a standard six-sided die and a fair coin. You roll first the die and get $1,2,3,4,5,6$ with equal prob. and then throw the coin and get $H,T$ with equal prob.
Define $A$ to be "get a $1$ with the die" and $B$ "get a $Head$ with the coin".
Clearly $P(A) = 1/6$ and $P(B) = 1/2$ and both event are independent. Therefore
$$
frac{P(A cap B)}{P(A)} = frac{1/12}{1/6} neq frac{1/12}{1/2} = frac{P(A cap B)}{P(B)}
$$
answered Oct 3 '16 at 22:32
ZubzubZubzub
3,8891922
$endgroup$
Say you have a standard six-sided die and a fair coin. You roll first the die and get $1,2,3,4,5,6$ with equal prob. and then throw the coin and get $H,T$ with equal prob.
Define $A$ to be "get a $1$ with the die" and $B$ "get a $Head$ with the coin".
Clearly $P(A) = 1/6$ and $P(B) = 1/2$ and both event are independent. Therefore
$$
frac{P(A cap B)}{P(A)} = frac{1/12}{1/6} neq frac{1/12}{1/2} = frac{P(A cap B)}{P(B)}
$$
answered Oct 3 '16 at 22:32
ZubzubZubzub
3,8891922
answered Oct 3 '16 at 22:32
ZubzubZubzub
3,8891922
answered Oct 3 '16 at 22:32
ZubzubZubzub
3,8891922
answered Oct 3 '16 at 22:32
ZubzubZubzub
3,8891922
3,8891922
add a comment |
add a comment |
$begingroup$
Since
$$P(Acap B)=P(A)P(B|A)=P(B)P(A|B)$$
We can see
$$frac{P(Acap B)}{P(A)}=frac{P(A)P(B|A)}{P(A)}=P(B|A)$$
but
$$frac{P(Acap B)}{P(B)}=frac{P(B)P(A|B)}{P(B)}=P(A|B)$$
Hence,
$$frac{P(Acap B)}{P(B)}neqfrac{P(Acap B)}{P(A)}$$
The example is when $A$ and $B$ are independent. In such case $P(A|B)=P(A)$ but $P(B|A)=P(B)$
answered Oct 3 '16 at 22:33
msmmsm
6,2292829
$endgroup$
add a comment |
$begingroup$
Since
$$P(Acap B)=P(A)P(B|A)=P(B)P(A|B)$$
We can see
$$frac{P(Acap B)}{P(A)}=frac{P(A)P(B|A)}{P(A)}=P(B|A)$$
but
$$frac{P(Acap B)}{P(B)}=frac{P(B)P(A|B)}{P(B)}=P(A|B)$$
Hence,
$$frac{P(Acap B)}{P(B)}neqfrac{P(Acap B)}{P(A)}$$
The example is when $A$ and $B$ are independent. In such case $P(A|B)=P(A)$ but $P(B|A)=P(B)$
answered Oct 3 '16 at 22:33
msmmsm
6,2292829
$endgroup$
add a comment |
$begingroup$
Since
$$P(Acap B)=P(A)P(B|A)=P(B)P(A|B)$$
We can see
$$frac{P(Acap B)}{P(A)}=frac{P(A)P(B|A)}{P(A)}=P(B|A)$$
but
$$frac{P(Acap B)}{P(B)}=frac{P(B)P(A|B)}{P(B)}=P(A|B)$$
Hence,
$$frac{P(Acap B)}{P(B)}neqfrac{P(Acap B)}{P(A)}$$
The example is when $A$ and $B$ are independent. In such case $P(A|B)=P(A)$ but $P(B|A)=P(B)$
answered Oct 3 '16 at 22:33
msmmsm
6,2292829
$endgroup$
Since
$$P(Acap B)=P(A)P(B|A)=P(B)P(A|B)$$
We can see
$$frac{P(Acap B)}{P(A)}=frac{P(A)P(B|A)}{P(A)}=P(B|A)$$
but
$$frac{P(Acap B)}{P(B)}=frac{P(B)P(A|B)}{P(B)}=P(A|B)$$
Hence,
$$frac{P(Acap B)}{P(B)}neqfrac{P(Acap B)}{P(A)}$$
The example is when $A$ and $B$ are independent. In such case $P(A|B)=P(A)$ but $P(B|A)=P(B)$
answered Oct 3 '16 at 22:33
msmmsm
6,2292829
edited Oct 3 '16 at 22:40
answered Oct 3 '16 at 22:33
msmmsm
6,2292829
answered Oct 3 '16 at 22:33
msmmsm
6,2292829
answered Oct 3 '16 at 22:33
msmmsm
6,2292829
6,2292829
add a comment |
add a comment |
$begingroup$
Honestly, i think it does not(i might be wrong, but here we go).
Notice that $frac{P(A cap B)}{P(A)} $ is actually the probability of the event $B$ knowing that the event $A$ happened. (Here, your sample space is actually the event $A$, so this formula is nothing but $P(C) = frac{C}{S}$ where $C$ is an event and $S$ the sample space.
Now, think about it:
Is the probability of an event $A$, knowing that $B$ has happened, equal to the probability of an event $B$, knowing that $A$ happened? The answer is no when $P(A)$ is not equal to $P(B)$. And that can be easily demonstrated:
You have:
$$ frac{P(A cap B)}{P(A)} = frac{P(A cap B)}{P(B)} $$
Obviously, this equality only holds if
$$P(A) = P(B) $$.
An example:
I have two dices, one black an one white. Let A be the event of the black resulting in 6 and B the event of the white resulting in 6.
I throw the first dice, so i have:
$$P(A) = 1/6 $$
And then:
$$P(B) = 1/6 $$
Since both of the events are independent, we have that
$$P(A cap B) = P(A)P(B) $$
So:
$$P(A/B) = P(A) = P(B/A) = P(B) $$.
This is an example that the equality holds for $P(A) = P(B)$. Any case that this doesn't apply, the equality doesn't apply.
answered Oct 3 '16 at 22:42
Vitor C GoergenVitor C Goergen
8261416
$endgroup$
add a comment |
$begingroup$
Honestly, i think it does not(i might be wrong, but here we go).
Notice that $frac{P(A cap B)}{P(A)} $ is actually the probability of the event $B$ knowing that the event $A$ happened. (Here, your sample space is actually the event $A$, so this formula is nothing but $P(C) = frac{C}{S}$ where $C$ is an event and $S$ the sample space.
Now, think about it:
Is the probability of an event $A$, knowing that $B$ has happened, equal to the probability of an event $B$, knowing that $A$ happened? The answer is no when $P(A)$ is not equal to $P(B)$. And that can be easily demonstrated:
You have:
$$ frac{P(A cap B)}{P(A)} = frac{P(A cap B)}{P(B)} $$
Obviously, this equality only holds if
$$P(A) = P(B) $$.
An example:
I have two dices, one black an one white. Let A be the event of the black resulting in 6 and B the event of the white resulting in 6.
I throw the first dice, so i have:
$$P(A) = 1/6 $$
And then:
$$P(B) = 1/6 $$
Since both of the events are independent, we have that
$$P(A cap B) = P(A)P(B) $$
So:
$$P(A/B) = P(A) = P(B/A) = P(B) $$.
This is an example that the equality holds for $P(A) = P(B)$. Any case that this doesn't apply, the equality doesn't apply.
answered Oct 3 '16 at 22:42
Vitor C GoergenVitor C Goergen
8261416
$endgroup$
add a comment |
$begingroup$
Honestly, i think it does not(i might be wrong, but here we go).
Notice that $frac{P(A cap B)}{P(A)} $ is actually the probability of the event $B$ knowing that the event $A$ happened. (Here, your sample space is actually the event $A$, so this formula is nothing but $P(C) = frac{C}{S}$ where $C$ is an event and $S$ the sample space.
Now, think about it:
Is the probability of an event $A$, knowing that $B$ has happened, equal to the probability of an event $B$, knowing that $A$ happened? The answer is no when $P(A)$ is not equal to $P(B)$. And that can be easily demonstrated:
You have:
$$ frac{P(A cap B)}{P(A)} = frac{P(A cap B)}{P(B)} $$
Obviously, this equality only holds if
$$P(A) = P(B) $$.
An example:
I have two dices, one black an one white. Let A be the event of the black resulting in 6 and B the event of the white resulting in 6.
I throw the first dice, so i have:
$$P(A) = 1/6 $$
And then:
$$P(B) = 1/6 $$
Since both of the events are independent, we have that
$$P(A cap B) = P(A)P(B) $$
So:
$$P(A/B) = P(A) = P(B/A) = P(B) $$.
This is an example that the equality holds for $P(A) = P(B)$. Any case that this doesn't apply, the equality doesn't apply.
answered Oct 3 '16 at 22:42
Vitor C GoergenVitor C Goergen
8261416
$endgroup$
Honestly, i think it does not(i might be wrong, but here we go).
Notice that $frac{P(A cap B)}{P(A)} $ is actually the probability of the event $B$ knowing that the event $A$ happened. (Here, your sample space is actually the event $A$, so this formula is nothing but $P(C) = frac{C}{S}$ where $C$ is an event and $S$ the sample space.
Now, think about it:
Is the probability of an event $A$, knowing that $B$ has happened, equal to the probability of an event $B$, knowing that $A$ happened? The answer is no when $P(A)$ is not equal to $P(B)$. And that can be easily demonstrated:
You have:
$$ frac{P(A cap B)}{P(A)} = frac{P(A cap B)}{P(B)} $$
Obviously, this equality only holds if
$$P(A) = P(B) $$.
An example:
I have two dices, one black an one white. Let A be the event of the black resulting in 6 and B the event of the white resulting in 6.
I throw the first dice, so i have:
$$P(A) = 1/6 $$
And then:
$$P(B) = 1/6 $$
Since both of the events are independent, we have that
$$P(A cap B) = P(A)P(B) $$
So:
$$P(A/B) = P(A) = P(B/A) = P(B) $$.
This is an example that the equality holds for $P(A) = P(B)$. Any case that this doesn't apply, the equality doesn't apply.
answered Oct 3 '16 at 22:42
Vitor C GoergenVitor C Goergen
8261416
answered Oct 3 '16 at 22:42
Vitor C GoergenVitor C Goergen
8261416
answered Oct 3 '16 at 22:42
Vitor C GoergenVitor C Goergen
8261416
answered Oct 3 '16 at 22:42
Vitor C GoergenVitor C Goergen
8261416
8261416
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$begingroup$
Did you mean to switch $B$ and $A$ in the second fraction? If not, just divide both sides by $P(Acap B)$ and find some $P(A)ne P(B)$
$endgroup$
– Simply Beautiful Art
Oct 3 '16 at 22:28
$begingroup$
@SimpleArt Sorry, is the equation I wrote equivalent to $P(A|B) = P(B|A)$?
$endgroup$
– Yuna Kun
Oct 3 '16 at 22:32
$begingroup$
what if P(A) = 1 and P(B) = 1/2 and we assume independence - then you get $P(A cap B) = 1/2$ so LHS = 1/2 and RHS = 1
$endgroup$
– Cato
Oct 4 '16 at 9:38
$begingroup$
A way of explaining it to be not true is that for independent events the equation boils down to P(A) = P(B)
$endgroup$
– Cato
Oct 4 '16 at 9:43