Probability proof that $P(A|B)geq frac{2}{3}$
up vote
2
down vote
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Prove that if $P(A)=P(B)=frac{3}{4}$ then $P(A|B)geq frac{2}{3}$.
I know that $P(A|B)=frac{P(Acap B)}{P(B)}$, but I can't get problem this any further. Where should I start?
probability
edited Nov 14 at 13:53
greedoid
34.3k114488
asked Nov 14 at 13:28
jte
153
add a comment |
up vote
2
down vote
favorite
Prove that if $P(A)=P(B)=frac{3}{4}$ then $P(A|B)geq frac{2}{3}$.
I know that $P(A|B)=frac{P(Acap B)}{P(B)}$, but I can't get problem this any further. Where should I start?
probability
edited Nov 14 at 13:53
greedoid
34.3k114488
asked Nov 14 at 13:28
jte
153
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Prove that if $P(A)=P(B)=frac{3}{4}$ then $P(A|B)geq frac{2}{3}$.
I know that $P(A|B)=frac{P(Acap B)}{P(B)}$, but I can't get problem this any further. Where should I start?
probability
edited Nov 14 at 13:53
greedoid
34.3k114488
asked Nov 14 at 13:28
jte
153
Prove that if $P(A)=P(B)=frac{3}{4}$ then $P(A|B)geq frac{2}{3}$.
I know that $P(A|B)=frac{P(Acap B)}{P(B)}$, but I can't get problem this any further. Where should I start?
probability
probability
edited Nov 14 at 13:53
greedoid
34.3k114488
asked Nov 14 at 13:28
jte
153
edited Nov 14 at 13:53
greedoid
34.3k114488
asked Nov 14 at 13:28
jte
153
edited Nov 14 at 13:53
greedoid
34.3k114488
edited Nov 14 at 13:53
greedoid
34.3k114488
edited Nov 14 at 13:53
greedoid
34.3k114488
34.3k114488
asked Nov 14 at 13:28
jte
153
asked Nov 14 at 13:28
jte
153
asked Nov 14 at 13:28
jte
153
153
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
7
down vote
accepted
Remeber that $P(...)leq 1$, so: $$frac{P(Acap B)}{P(B)} ={P(A)+P(B)-P(Acup B)over P(B)}geq {2cdot {3over 4}-1over {3over 4}}={2over 3}$$
answered Nov 14 at 13:31
greedoid
34.3k114488
Ahh of course... simple but effective.
– jte
Nov 14 at 13:36
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
Remeber that $P(...)leq 1$, so: $$frac{P(Acap B)}{P(B)} ={P(A)+P(B)-P(Acup B)over P(B)}geq {2cdot {3over 4}-1over {3over 4}}={2over 3}$$
answered Nov 14 at 13:31
greedoid
34.3k114488
Ahh of course... simple but effective.
– jte
Nov 14 at 13:36
add a comment |
up vote
7
down vote
accepted
Remeber that $P(...)leq 1$, so: $$frac{P(Acap B)}{P(B)} ={P(A)+P(B)-P(Acup B)over P(B)}geq {2cdot {3over 4}-1over {3over 4}}={2over 3}$$
answered Nov 14 at 13:31
greedoid
34.3k114488
Ahh of course... simple but effective.
– jte
Nov 14 at 13:36
add a comment |
up vote
7
down vote
accepted
up vote
7
down vote
accepted
Remeber that $P(...)leq 1$, so: $$frac{P(Acap B)}{P(B)} ={P(A)+P(B)-P(Acup B)over P(B)}geq {2cdot {3over 4}-1over {3over 4}}={2over 3}$$
answered Nov 14 at 13:31
greedoid
34.3k114488
Remeber that $P(...)leq 1$, so: $$frac{P(Acap B)}{P(B)} ={P(A)+P(B)-P(Acup B)over P(B)}geq {2cdot {3over 4}-1over {3over 4}}={2over 3}$$
answered Nov 14 at 13:31
greedoid
34.3k114488
answered Nov 14 at 13:31
greedoid
34.3k114488
answered Nov 14 at 13:31
greedoid
34.3k114488
answered Nov 14 at 13:31
greedoid
34.3k114488
34.3k114488
Ahh of course... simple but effective.
– jte
Nov 14 at 13:36
add a comment |
Ahh of course... simple but effective.
– jte
Nov 14 at 13:36
Ahh of course... simple but effective.
– jte
Nov 14 at 13:36
Ahh of course... simple but effective.
– jte
Nov 14 at 13:36
add a comment |
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