Question about joint probability

[theone]

Homework Statement

Let N denote the number of accidents occurring during one month on the northbound side of a highway and let S denote the number occurring on the southbound side. Suppose that N and S are jointly distributed as indicated in the table.

N/S 0 1 2 3+
0 0.04 0.06 0.10 0.04
1 0.10 0.18 0.08 0.03
2 0.12 0.06 0.05 0.02
3+ 0.05 0.04 0.02 0.0

Find P(N=1|N+S=2)

Homework Equations

The Attempt at a Solution

P(N=1 | N+S is not equal to 2, as in the table) = 0.10+0.18+0.08+0.03 = 0.39

I know that P(N+S) = 2 is the sum of all the different possibilities in the table where N+S=2, so 0.18 + 0.12 + 0.10 = 0.4

I do not know how to put this information together to answer the question : P(N=1 | N+S=2)

 

[PeroK]

$P(N = 1| N +S = 2)$ means the probability that $N = 1$ given that $N+S = 2$.

So, if $N+S = 2$ what is the probability that $N = 1$?

Or, using the frequency approach, from all the times that $N+S = 2$, how often is $N = 1$?

By the way, I think the first entry in your table should be $0.04$.

 

[theone]

Or, using the frequency approach, from all the times that $N+S = 2$, how often is $N = 1$?

would that be 1/3?

In the solution, I am seeing that the way to get P(N=1 | N+S=2) is to do P(N=1 | S=1)/P(N+S=2) = 0.18/0.40 = 0.45
but I do not understand the reason for dividing the probabilities

and the first number should be 0.04, I will edit it

 

[PeroK]

would that be 1/3?

In the solution, I am seeing that the way to get P(N=1 | N+S=2) is to do P(N=1 | S=1)/P(N+S=2) = 0.18/0.40 = 0.45
but I do not understand the reason for dividing the probabilities

and the first number should be 0.04, I will edit it

I like the frequentist approach. So, let's take 100 typical days and see what happens:

1) On 40 of the days, $N + S = 2$.

1.1) On 18 of these days, $N = 1$

1.2) On 22 of these days, $N \ne 1$

2) On 60 of the days, $N + S \ne 2$

2.1) On 21 of those days, $N = 1$.

2.2) On 39 of these days, $N \ne 1$

In terms of these two factors, that is a complete breakdown of the numbers.

Now, when you have a conditional probability, such as in this case "given $N + S =2$", then you are restricting your interest to those cases. Formally, this becomes your new "sample space".

In this case, we are only interest in 40 days out of every 100. (The other days we ignore. They do not matter.) What happens on those 40 days that we are interested in?

1) On 18 days out of the 40, we have $N = 1$, that's 45% of the time.

2) On 22 days out of the 40, we have $N \ne 1$, that's 55% of the time.

Therefore:

$P(N=1|N + S =2) = 0.45$

Does that all make sense?

 

[Ray Vickson]

Homework Statement

Let N denote the number of accidents occurring during one month on the northbound side of a highway and let S denote the number occurring on the southbound side. Suppose that N and S are jointly distributed as indicated in the table.

N/S 0 1 2 3+
0 0.04 0.06 0.10 0.04
1 0.10 0.18 0.08 0.03
2 0.12 0.06 0.05 0.02
3+ 0.05 0.04 0.02 0.0

Find P(N=1|N+S=2)

Homework Equations

The Attempt at a Solution

P(N=1 | N+S is not equal to 2, as in the table) = 0.10+0.18+0.08+0.03 = 0.39

I know that P(N+S) = 2 is the sum of all the different possibilities in the table where N+S=2, so 0.18 + 0.12 + 0.10 = 0.4

I do not know how to put this information together to answer the question : P(N=1 | N+S=2)

Just use the definition of conditional probability: $P(A|B) = P(A \& B)/P(B).$ So,
$$P(N=1|N+S=2) = \frac{P(N=1 \; \& \; N+S=2)}{P(N+S=2)}.$$ Can you write the event $\{ N=1 \: \& \: N+S=2 \}$ in a simpler way?

 

[theone]

I like the frequentist approach. So, let's take 100 typical days and see what happens:

1) On 40 of the days, $N + S = 2$.

1.1) On 18 of these days, $N = 1$

1.2) On 22 of these days, $N \ne 1$

2) On 60 of the days, $N + S \ne 2$

2.1) On 21 of those days, $N = 1$.

2.2) On 39 of these days, $N \ne 1$

In terms of these two factors, that is a complete breakdown of the numbers.

Now, when you have a conditional probability, such as in this case "given $N + S =2$", then you are restricting your interest to those cases. Formally, this becomes your new "sample space".

In this case, we are only interest in 40 days out of every 100. (The other days we ignore. They do not matter.) What happens on those 40 days that we are interested in?

1) On 18 days out of the 40, we have $N = 1$, that's 45% of the time.

2) On 22 days out of the 40, we have $N \ne 1$, that's 55% of the time.

Therefore:

$P(N=1|N + S =2) = 0.45$

Does that all make sense?

yes, it does. Thanks
So then the reason that you can't say P(N=1 | N+S=2) = (sum of the "1" row) * (sum of the "1" column) is because those two probabilities being multiplied are from a different "sample space", the one where N+S is not equal to 2 ?

Just use the definition of conditional probability: $P(A|B) = P(A \& B)/P(B).$ So,
$$P(N=1|N+S=2) = \frac{P(N=1 \; \& \; N+S=2)}{P(N+S=2)}.$$ Can you write the event $\{ N=1 \: \& \: N+S=2 \}$ in a simpler way?

Thanks, I think the problem is that it's hard for me to see N+S=2 and N=1 as two "events" A and B where A is conditional on B.
It feels very different than for example: A is the event of you picking a shirt from a bag of blue and red shirts, and B is the event that the shirt is a particular brand

 

[Ray Vickson]

yes, it does. Thanks
So then the reason that you can't say P(N=1 | N+S=2) = (sum of the "1" row) * (sum of the "1" column) is because those two probabilities being multiplied are from a different "sample space", the one where N+S is not equal to 2 ?

Thanks, I think the problem is that it's hard for me to see N+S=2 and N=1 as two "events" A and B where A is conditional on B.
It feels very different than for example: A is the event of you picking a shirt from a bag of blue and red shirts, and B is the event that the shirt is a particular brand

It is really very much like your shirts example, except that one attribute refers to accidents in the north lane and the other to accidents in the south lane (and for all we know, multi-vehicle accidents in both lanes at the same time).