Proving that if $A$ and $B$ are events, then $Pr[A ∩ B] ge Pr[A] + Pr[B] - 1$Probability of the union of $3$ events?$K$ events that are $(K-1)$-wise Independent but not Mutually/Fully IndependentProbability of a Union of EventsProving that three events are mutually independentProving that $Pleft ( bigcup_i=1^nA_i right )leq sum_i=1^nP(A_i)$ by inductionProbability: Prove that events are independentCan two different events generate the same set of probabilities?If $A$ and B are two events such that $P(A) > 0$ and $P(A) + P(B) > 1$Given $A, B$, and $C$ are mutually independent events, find $ P(A cap B' cap C')$Bounding the probability of a set of non independent events
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Proving that if $A$ and $B$ are events, then $Pr[A ∩ B] ge Pr[A] + Pr[B] - 1$
Probability of the union of $3$ events?$K$ events that are $(K-1)$-wise Independent but not Mutually/Fully IndependentProbability of a Union of EventsProving that three events are mutually independentProving that $Pleft ( bigcup_i=1^nA_i right )leq sum_i=1^nP(A_i)$ by inductionProbability: Prove that events are independentCan two different events generate the same set of probabilities?If $A$ and B are two events such that $P(A) > 0$ and $P(A) + P(B) > 1$Given $A, B$, and $C$ are mutually independent events, find $ P(A cap B' cap C')$Bounding the probability of a set of non independent events
$begingroup$
I am just looking for some guidance whether I am approaching the questions correctly. New to probability!
Prove that if $A$ and $B$ are events, then $$Pr[A cap B] geq Pr[A]+Pr[B]-1$$
I rewrote the statement in the following way :
$$1 geq Pr[A] + Pr[B] - Pr[A cap B]$$
And based on some probability properties, I rewrote out the statement again:
$$1 geq Pr[A cup B]$$
Is this enough to prove the statement?
probability proof-verification
edited Mar 20 at 19:55
Maria Mazur
49.5k1361124
asked Mar 20 at 17:49
RobinRobin
525
$endgroup$
add a comment |
$begingroup$
I am just looking for some guidance whether I am approaching the questions correctly. New to probability!
Prove that if $A$ and $B$ are events, then $$Pr[A cap B] geq Pr[A]+Pr[B]-1$$
I rewrote the statement in the following way :
$$1 geq Pr[A] + Pr[B] - Pr[A cap B]$$
And based on some probability properties, I rewrote out the statement again:
$$1 geq Pr[A cup B]$$
Is this enough to prove the statement?
probability proof-verification
edited Mar 20 at 19:55
Maria Mazur
49.5k1361124
asked Mar 20 at 17:49
RobinRobin
525
$endgroup$
2
$begingroup$
Perfect! ... ...
$endgroup$
– Maria Mazur
Mar 20 at 17:51
add a comment |
$begingroup$
I am just looking for some guidance whether I am approaching the questions correctly. New to probability!
Prove that if $A$ and $B$ are events, then $$Pr[A cap B] geq Pr[A]+Pr[B]-1$$
I rewrote the statement in the following way :
$$1 geq Pr[A] + Pr[B] - Pr[A cap B]$$
And based on some probability properties, I rewrote out the statement again:
$$1 geq Pr[A cup B]$$
Is this enough to prove the statement?
probability proof-verification
edited Mar 20 at 19:55
Maria Mazur
49.5k1361124
asked Mar 20 at 17:49
RobinRobin
525
$endgroup$
I am just looking for some guidance whether I am approaching the questions correctly. New to probability!
Prove that if $A$ and $B$ are events, then $$Pr[A cap B] geq Pr[A]+Pr[B]-1$$
I rewrote the statement in the following way :
$$1 geq Pr[A] + Pr[B] - Pr[A cap B]$$
And based on some probability properties, I rewrote out the statement again:
$$1 geq Pr[A cup B]$$
Is this enough to prove the statement?
probability proof-verification
probability proof-verification
edited Mar 20 at 19:55
Maria Mazur
49.5k1361124
asked Mar 20 at 17:49
RobinRobin
525
edited Mar 20 at 19:55
Maria Mazur
49.5k1361124
asked Mar 20 at 17:49
RobinRobin
525
edited Mar 20 at 19:55
Maria Mazur
49.5k1361124
edited Mar 20 at 19:55
Maria Mazur
49.5k1361124
edited Mar 20 at 19:55
Maria Mazur
49.5k1361124
49.5k1361124
asked Mar 20 at 17:49
RobinRobin
525
asked Mar 20 at 17:49
RobinRobin
525
asked Mar 20 at 17:49
RobinRobin
525
525
2
$begingroup$
Perfect! ... ...
$endgroup$
– Maria Mazur
Mar 20 at 17:51
add a comment |
2
$begingroup$
Perfect! ... ...
$endgroup$
– Maria Mazur
Mar 20 at 17:51
2
2
$begingroup$
Perfect! ... ...
$endgroup$
– Maria Mazur
Mar 20 at 17:51
$begingroup$
Perfect! ... ...
$endgroup$
– Maria Mazur
Mar 20 at 17:51
add a comment |
1 Answer
1
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$begingroup$
So you use $$P(Acup B ) = P(A)+P(B)-P(Acap B)$$ and that $P(Acup B)leq 1$
So you got $$1geq P(A)+P(B)-P(Acap B)$$ and now swap $1$ and $P(Acap B)$ and you got what you want.
That is correct.
answered Mar 20 at 17:52
Maria MazurMaria Mazur
49.5k1361124
$endgroup$
add a comment |
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$begingroup$
So you use $$P(Acup B ) = P(A)+P(B)-P(Acap B)$$ and that $P(Acup B)leq 1$
So you got $$1geq P(A)+P(B)-P(Acap B)$$ and now swap $1$ and $P(Acap B)$ and you got what you want.
That is correct.
answered Mar 20 at 17:52
Maria MazurMaria Mazur
49.5k1361124
$endgroup$
add a comment |
$begingroup$
So you use $$P(Acup B ) = P(A)+P(B)-P(Acap B)$$ and that $P(Acup B)leq 1$
So you got $$1geq P(A)+P(B)-P(Acap B)$$ and now swap $1$ and $P(Acap B)$ and you got what you want.
That is correct.
answered Mar 20 at 17:52
Maria MazurMaria Mazur
49.5k1361124
$endgroup$
add a comment |
$begingroup$
So you use $$P(Acup B ) = P(A)+P(B)-P(Acap B)$$ and that $P(Acup B)leq 1$
So you got $$1geq P(A)+P(B)-P(Acap B)$$ and now swap $1$ and $P(Acap B)$ and you got what you want.
That is correct.
answered Mar 20 at 17:52
Maria MazurMaria Mazur
49.5k1361124
$endgroup$
So you use $$P(Acup B ) = P(A)+P(B)-P(Acap B)$$ and that $P(Acup B)leq 1$
So you got $$1geq P(A)+P(B)-P(Acap B)$$ and now swap $1$ and $P(Acap B)$ and you got what you want.
That is correct.
answered Mar 20 at 17:52
Maria MazurMaria Mazur
49.5k1361124
answered Mar 20 at 17:52
Maria MazurMaria Mazur
49.5k1361124
answered Mar 20 at 17:52
Maria MazurMaria Mazur
49.5k1361124
answered Mar 20 at 17:52
Maria MazurMaria Mazur
49.5k1361124
49.5k1361124
add a comment |
add a comment |
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– Maria Mazur
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