Basic conditional probability problem
$begingroup$
I found this problem in the Sheldon Ross book:
In a certain community, 36 percent of the families
own a dog and 22 percent of the families that own
a dog also own a cat. In addition, 30 percent of the
families own a cat. What is
(a) the probability that a randomly selected family
owns both a dog and a cat?
(b) the conditional probability that a randomly
selected family owns a dog given that it owns
a cat?
The answer to a) is given as $$P(Dcap C) = P(Cvert D)P(D)$$
This is very non-intuitive to me as I can't understand why $$P(Dcap C) not = P(C vert D) not = P(D vert C)$$
conditional-probability
asked Dec 4 '18 at 1:18
Bindi CapriqiBindi Capriqi
1
$endgroup$
add a comment |
$begingroup$
I found this problem in the Sheldon Ross book:
In a certain community, 36 percent of the families
own a dog and 22 percent of the families that own
a dog also own a cat. In addition, 30 percent of the
families own a cat. What is
(a) the probability that a randomly selected family
owns both a dog and a cat?
(b) the conditional probability that a randomly
selected family owns a dog given that it owns
a cat?
The answer to a) is given as $$P(Dcap C) = P(Cvert D)P(D)$$
This is very non-intuitive to me as I can't understand why $$P(Dcap C) not = P(C vert D) not = P(D vert C)$$
conditional-probability
asked Dec 4 '18 at 1:18
Bindi CapriqiBindi Capriqi
1
$endgroup$
add a comment |
$begingroup$
I found this problem in the Sheldon Ross book:
In a certain community, 36 percent of the families
own a dog and 22 percent of the families that own
a dog also own a cat. In addition, 30 percent of the
families own a cat. What is
(a) the probability that a randomly selected family
owns both a dog and a cat?
(b) the conditional probability that a randomly
selected family owns a dog given that it owns
a cat?
The answer to a) is given as $$P(Dcap C) = P(Cvert D)P(D)$$
This is very non-intuitive to me as I can't understand why $$P(Dcap C) not = P(C vert D) not = P(D vert C)$$
conditional-probability
asked Dec 4 '18 at 1:18
Bindi CapriqiBindi Capriqi
1
$endgroup$
I found this problem in the Sheldon Ross book:
In a certain community, 36 percent of the families
own a dog and 22 percent of the families that own
a dog also own a cat. In addition, 30 percent of the
families own a cat. What is
(a) the probability that a randomly selected family
owns both a dog and a cat?
(b) the conditional probability that a randomly
selected family owns a dog given that it owns
a cat?
The answer to a) is given as $$P(Dcap C) = P(Cvert D)P(D)$$
This is very non-intuitive to me as I can't understand why $$P(Dcap C) not = P(C vert D) not = P(D vert C)$$
conditional-probability
conditional-probability
asked Dec 4 '18 at 1:18
Bindi CapriqiBindi Capriqi
1
asked Dec 4 '18 at 1:18
Bindi CapriqiBindi Capriqi
1
asked Dec 4 '18 at 1:18
Bindi CapriqiBindi Capriqi
1
asked Dec 4 '18 at 1:18
Bindi CapriqiBindi Capriqi
1
asked Dec 4 '18 at 1:18
Bindi CapriqiBindi Capriqi
1
1
add a comment |
add a comment |
1 Answer
1
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votes
$begingroup$
You are someone who always bring an umbrella whenever it rains, also, sometimes you might bring an umbrella even when it is not raining.
Let $D$ be the event that it rains, and $C$ is the event that you bring an umbrella.
$$P(C|D)=1$$
However, it doesn't means that it rains all the time, $P(C cap D)<1$.
Also, it is unlikely that whenever you bring an umbrella, it brings rain. $P(D|C)<1$.
answered Dec 4 '18 at 1:33
Siong Thye GohSiong Thye Goh
100k1466117
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add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
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active
oldest
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$begingroup$
You are someone who always bring an umbrella whenever it rains, also, sometimes you might bring an umbrella even when it is not raining.
Let $D$ be the event that it rains, and $C$ is the event that you bring an umbrella.
$$P(C|D)=1$$
However, it doesn't means that it rains all the time, $P(C cap D)<1$.
Also, it is unlikely that whenever you bring an umbrella, it brings rain. $P(D|C)<1$.
answered Dec 4 '18 at 1:33
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
add a comment |
$begingroup$
You are someone who always bring an umbrella whenever it rains, also, sometimes you might bring an umbrella even when it is not raining.
Let $D$ be the event that it rains, and $C$ is the event that you bring an umbrella.
$$P(C|D)=1$$
However, it doesn't means that it rains all the time, $P(C cap D)<1$.
Also, it is unlikely that whenever you bring an umbrella, it brings rain. $P(D|C)<1$.
answered Dec 4 '18 at 1:33
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
add a comment |
$begingroup$
You are someone who always bring an umbrella whenever it rains, also, sometimes you might bring an umbrella even when it is not raining.
Let $D$ be the event that it rains, and $C$ is the event that you bring an umbrella.
$$P(C|D)=1$$
However, it doesn't means that it rains all the time, $P(C cap D)<1$.
Also, it is unlikely that whenever you bring an umbrella, it brings rain. $P(D|C)<1$.
answered Dec 4 '18 at 1:33
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
You are someone who always bring an umbrella whenever it rains, also, sometimes you might bring an umbrella even when it is not raining.
Let $D$ be the event that it rains, and $C$ is the event that you bring an umbrella.
$$P(C|D)=1$$
However, it doesn't means that it rains all the time, $P(C cap D)<1$.
Also, it is unlikely that whenever you bring an umbrella, it brings rain. $P(D|C)<1$.
answered Dec 4 '18 at 1:33
Siong Thye GohSiong Thye Goh
100k1466117
answered Dec 4 '18 at 1:33
Siong Thye GohSiong Thye Goh
100k1466117
answered Dec 4 '18 at 1:33
Siong Thye GohSiong Thye Goh
100k1466117
answered Dec 4 '18 at 1:33
Siong Thye GohSiong Thye Goh
100k1466117
100k1466117
add a comment |
add a comment |
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