1. ## Conditional Probability

Events , and form a partiton of the sample space S with probabilities , , .
If E is an event in S with , ,

Im supposed to find
P(E)=
P(A1 and E)

2. Originally Posted by mjgtheg
Events , and form a partiton of the sample space S with probabilities , , .
If E is an event in S with , ,

Im supposed to find
P(E)=
P(A1 and E)

P(E)

Ok, there are 3 ways in which you can get E.
You can get A1 and then E.
You can get A2 and then E.
Or You can get A3 and then E.
Any of these will give you E.

So.. What is the probability of the first one? Well, the chances of getting A1 are 0.4 (given in the question: P(A1)=0.4) Then the chance of getting an E on top of that is also 0.4. (also given: P(E|A1)=0.4).
So the chance of getting an A1 and then an E is 0.4 * 0.4 = 0.16.

Similarly, you can work out the probability of getting an E one of the other two ways.
Then to get the total probability of getting an E, you simply add up the probabilities of each possible way of ending up with an E.

.........

By P(A1 and E) do you mean P(A1 U E)? As in, the probability of getting either an A, or an E, or both?
Because the probability of getting A1 and then E we have calculated in the previous part of the question...

If you mean U, it's easy to calculate, similarly to the first part - just be careful not to count P(E|A1) twice!

3. I filled in part 1.....

Originally Posted by mjgtheg
Events , and form a partiton of the sample space S with probabilities , , .
If E is an event in S with , ,

P(E)=P(E|A1)P(A1)+P(E|A2)P(A2)+P(E|A3)P(A3)

4. Hello,

Since $\displaystyle A_1,A_2,A_3$ form a partition of the sample space, we have :

$\displaystyle P(E)=P(E \cap (A_1 \cup A_2 \cup A_3))=P(E \cap A_1)+P(E \cap A_2)+P(E\cap A_3)$
then use the definition of conditional probability and you're done...