# Math Help - Conditional probability

1. ## Conditional probability

I have a sequence of states (either 1 or 2).

For eg: Out of 100, if 30 of them are '1' and 70 of them are '2' then the P(1)=.3 and P(2)=.7

Now I need to find the probability of occurance of 1 in the next state when the current state is already 1 P_(11). Also find the probability of occurance of 2 when the current state is 1 (P_12).

2. ## Re: Conditional probability

Originally Posted by lemontree45
I have a sequence of states (either 1 or 2).

For eg: Out of 100, if 30 of them are '1' and 70 of them are '2' then the P(1)=.3 and P(2)=.7

Now I need to find the probability of occurance of 1 in the next state when the current state is already 1 P_(11). Also find the probability of occurance of 2 when the current state is 1 (P_12).
Is that example the actual probabilities you were given, or were you just trying to illustrate that the states 1 and 2 are exhaustive? Is it fair to assume that the next state is independent of the previous state? What is it about this problem specifically that you are having trouble with?

3. ## Re: Conditional probability

Hello, lemontree45!

I have a sequence of states (either 1 or 2).

For eg: Out of 100, if 30 of them are '1' and 70 of them are '2',
then the P(1) = 0.3 and P(2) = 0.7

Now I need to find the probability of occurance of 1 in the next state
when the current state is already 1: P(11).
Also find the probability of occurance of 2 when the current state is 1: P(12).

I will ignore the reference to "100" and assume independent probabilities.
. . That is, P(1) is always 0.3 . . . and P(2) is always 0.7.

Then we have:

. . $\begin{Bmatrix}P(11) \:=\:(0.3)(0.3) \:=\:0.09 && P(12) \:=\:(0.3)(0.7) \:=\:0.21 \\ \\[-3mm] P(21) \:=\:(0.7)(0.3) \:=\:0.21 && P(22) \:=\:(0.7)(0.7) \:=\:0.49 \end{Bmatrix}$