# Homework Help: Probability question

1. Jul 5, 2005

### Townsend

If event S has a 40 percent chance of happening on the first try and a 65 percent chance of happening on the second try what is the chance of event A happening?

Let A be the event S happens on the first try, note $$A^c$$ is the complement of A.
Let B be the event S happens on the second try.

I reasoned it to be a 79 percent chance of S happening.

1. A and B are mutually exclusive, in other words if A happens B does not, vise versa.

2. The probability of event B happening depends on event A not happening.

Thus $$P(B \cap A^c) = .39$$, so P(S) = P(A) + $$P(B \cap A^c)$$ = .79.

So I reason a 79 percent chance of event S.

Thanks

2. Jul 5, 2005

### honestrosewater

You mean of event S happening?
Why do you think they are mutually exclusive? Does the problem say so? Why do you think they are dependent?

The first paragraph asks for the probability that, in two tries, S occurs. That would mean either S occurs on the first try or S occurs on the second try or S occurs on both tries. Think of getting heads on two coin tosses. Either the first toss is heads or the second toss is heads or both tosses are heads. What formula do you use for these "or" situations?

3. Jul 5, 2005

### Townsend

Yes, sorry.

I didn't state the problem as it was stated in the text to save some typing. The problem is

A salesman has a 40 percent chance of making a sale the first time he visits a house and a 65 percent chance of making a sale the second time he visits that same house and he will only ever visit a house twice. What is the chance he will make a sale?

Based on this I can only reason that he will either make a sale the first day or second day but not both days. So I suppose that makes the two events mutually exclusive.

4. Jul 5, 2005

### honestrosewater

Yeah, that's not the greatest. Maybe he's selling different things each time, there could be different people living in the house, etc. Go with whatever you're willing to defend. I would go with the most straightforward- A and B are non-mutually exclusive, independent events. You could also solve it for each scenario. Do you have the formula for P(A or B) (or P(A $\cup$ B))?

Last edited: Jul 5, 2005
5. Jul 5, 2005

### Townsend

Not sure... If A is an event and B is an event then the sample points in A or B would just be A union B, right? The problem is figuring out what those sample points are. If you accept that B is dependent on A and A intersect B is null then how do you find P(A or B) if you're only given what P(A) and P(B) are, and P(B) is given with the assumption that A will not happen?

6. Jul 5, 2005

### honestrosewater

Right, P(A or B) is the same as P(A U B). P(A U B) = P(A) + P(B) - P(A $\cap$ B).
If A and B are mutually exclusive, what is P(A $\cap$ B)?
If A and B are independent, what is P(A $\cap$ B)?
So why would you think A and B are dependent (but not mutually exclusive)? In your original post, you said you thought A and B were mutually exclusive. If they are, the probability of A occurring given that B has occurred is what? The probability of B occurring given that A has occurred is what?

The reason I would go with A and B being independent is that, by default, you don't assume more than the problem tells you. If A and B were mutually exclusive, the problem should have added:
"But if he makes a sale the first try, he doesn't make a sale the second try, and if he makes a sale the second try...." or something similar.
If A and B were dependent, the problem should have added:
"If he makes a sale the first try, then on the second try..."
or something else to tell you how the events depend on each other. It doesn't say anything about the events being dependent or mutually exclusive, and there's nothing inherent in the situation that makes the events dependent or mutually exclusive, so you should just assume the events are independent. You could argue that a salesman wouldn't even try to make a sale on the second try if he made a sale on the first try, but you should be prepared to really make that argument.

Last edited: Jul 5, 2005
7. Jul 5, 2005