Mixing
Conditional probability of continuous random variable
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I understand that the formula for calculating a conditional probability is the following $P(A mid B) = frac{P(A cap B)}{P(A)}$
I have this probability to calculate: $P(2le X le 3 mid X ge1)$.
What I would normally do is $P(2le X le 3 mid X ge1) = frac{P(2le X le 3 cap X ge1)}{P(2le X le 3)}$.
However the numerator equals $P(2le X le 3)$. The above probability automatically becomes $1$ which is not right.
The textbook says the correct formula is this: $P(2le X le 3 mid X ge1) = frac{P(2le X le 3 cap X ge1)}{P(X ge1)}$.
Why is that? I thought the denominator is always the first part of the conditional probability like the first formula.
probability random-variables conditional-probability
asked Jan 16 at 22:00
Harry TouloupasHarry Touloupas
124
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add a comment |
$begingroup$
I understand that the formula for calculating a conditional probability is the following $P(A mid B) = frac{P(A cap B)}{P(A)}$
I have this probability to calculate: $P(2le X le 3 mid X ge1)$.
What I would normally do is $P(2le X le 3 mid X ge1) = frac{P(2le X le 3 cap X ge1)}{P(2le X le 3)}$.
However the numerator equals $P(2le X le 3)$. The above probability automatically becomes $1$ which is not right.
The textbook says the correct formula is this: $P(2le X le 3 mid X ge1) = frac{P(2le X le 3 cap X ge1)}{P(X ge1)}$.
Why is that? I thought the denominator is always the first part of the conditional probability like the first formula.
probability random-variables conditional-probability
asked Jan 16 at 22:00
Harry TouloupasHarry Touloupas
124
$endgroup$
1
$begingroup$
en.wikipedia.org/wiki/Bayes%27_theorem#Statement_of_theorem
$endgroup$
– d.k.o.
Jan 16 at 22:04
add a comment |
$begingroup$
I understand that the formula for calculating a conditional probability is the following $P(A mid B) = frac{P(A cap B)}{P(A)}$
I have this probability to calculate: $P(2le X le 3 mid X ge1)$.
What I would normally do is $P(2le X le 3 mid X ge1) = frac{P(2le X le 3 cap X ge1)}{P(2le X le 3)}$.
However the numerator equals $P(2le X le 3)$. The above probability automatically becomes $1$ which is not right.
The textbook says the correct formula is this: $P(2le X le 3 mid X ge1) = frac{P(2le X le 3 cap X ge1)}{P(X ge1)}$.
Why is that? I thought the denominator is always the first part of the conditional probability like the first formula.
probability random-variables conditional-probability
asked Jan 16 at 22:00
Harry TouloupasHarry Touloupas
124
$endgroup$
I understand that the formula for calculating a conditional probability is the following $P(A mid B) = frac{P(A cap B)}{P(A)}$
I have this probability to calculate: $P(2le X le 3 mid X ge1)$.
What I would normally do is $P(2le X le 3 mid X ge1) = frac{P(2le X le 3 cap X ge1)}{P(2le X le 3)}$.
However the numerator equals $P(2le X le 3)$. The above probability automatically becomes $1$ which is not right.
The textbook says the correct formula is this: $P(2le X le 3 mid X ge1) = frac{P(2le X le 3 cap X ge1)}{P(X ge1)}$.
Why is that? I thought the denominator is always the first part of the conditional probability like the first formula.
probability random-variables conditional-probability
probability random-variables conditional-probability
asked Jan 16 at 22:00
Harry TouloupasHarry Touloupas
124
asked Jan 16 at 22:00
Harry TouloupasHarry Touloupas
124
asked Jan 16 at 22:00
Harry TouloupasHarry Touloupas
124
asked Jan 16 at 22:00
Harry TouloupasHarry Touloupas
124
asked Jan 16 at 22:00
Harry TouloupasHarry Touloupas
124
124
1
$begingroup$
en.wikipedia.org/wiki/Bayes%27_theorem#Statement_of_theorem
$endgroup$
– d.k.o.
Jan 16 at 22:04
add a comment |
1
$begingroup$
en.wikipedia.org/wiki/Bayes%27_theorem#Statement_of_theorem
$endgroup$
– d.k.o.
Jan 16 at 22:04
1
1
$begingroup$
en.wikipedia.org/wiki/Bayes%27_theorem#Statement_of_theorem
$endgroup$
– d.k.o.
Jan 16 at 22:04
$begingroup$
en.wikipedia.org/wiki/Bayes%27_theorem#Statement_of_theorem
$endgroup$
– d.k.o.
Jan 16 at 22:04
add a comment |
1 Answer
1
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oldest
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$begingroup$
The formula for calculating a conditional probability that you wrote is not correct; you should divide by $P(B)$. See https://en.wikipedia.org/wiki/Conditional_probability
answered Jan 16 at 22:05
user144410user144410
1,0282719
$endgroup$
$begingroup$
You are right, professor's notes were wrong. Thank you.
$endgroup$
– Harry Touloupas
Jan 16 at 22:08
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The formula for calculating a conditional probability that you wrote is not correct; you should divide by $P(B)$. See https://en.wikipedia.org/wiki/Conditional_probability
answered Jan 16 at 22:05
user144410user144410
1,0282719
$endgroup$
$begingroup$
You are right, professor's notes were wrong. Thank you.
$endgroup$
– Harry Touloupas
Jan 16 at 22:08
add a comment |
$begingroup$
The formula for calculating a conditional probability that you wrote is not correct; you should divide by $P(B)$. See https://en.wikipedia.org/wiki/Conditional_probability
answered Jan 16 at 22:05
user144410user144410
1,0282719
$endgroup$
$begingroup$
You are right, professor's notes were wrong. Thank you.
$endgroup$
– Harry Touloupas
Jan 16 at 22:08
add a comment |
$begingroup$
The formula for calculating a conditional probability that you wrote is not correct; you should divide by $P(B)$. See https://en.wikipedia.org/wiki/Conditional_probability
answered Jan 16 at 22:05
user144410user144410
1,0282719
$endgroup$
The formula for calculating a conditional probability that you wrote is not correct; you should divide by $P(B)$. See https://en.wikipedia.org/wiki/Conditional_probability
answered Jan 16 at 22:05
user144410user144410
1,0282719
answered Jan 16 at 22:05
user144410user144410
1,0282719
answered Jan 16 at 22:05
user144410user144410
1,0282719
answered Jan 16 at 22:05
user144410user144410
1,0282719
1,0282719
$begingroup$
You are right, professor's notes were wrong. Thank you.
$endgroup$
– Harry Touloupas
Jan 16 at 22:08
add a comment |
$begingroup$
You are right, professor's notes were wrong. Thank you.
$endgroup$
– Harry Touloupas
Jan 16 at 22:08
$begingroup$
You are right, professor's notes were wrong. Thank you.
$endgroup$
– Harry Touloupas
Jan 16 at 22:08
$begingroup$
You are right, professor's notes were wrong. Thank you.
$endgroup$
– Harry Touloupas
Jan 16 at 22:08
add a comment |
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$begingroup$
en.wikipedia.org/wiki/Bayes%27_theorem#Statement_of_theorem
$endgroup$
– d.k.o.
Jan 16 at 22:04