Solving a Probability Problem: Get Help Here

[Nex Vortex]

I'm trying to figure out the probability of getting something tomorrow, but since I've never taken a statistics course, I don't really know how to do a problem of this nature.

There are two events, I will call the macro-events, M and N for argument sake, each with a 50% chance of happening. In each macro-event, there are also three micro-events, I will call M₁, M₂, M₃, N₁, etc. Each micro-event has a 20% chance of a desired result. After a desired result is reached, it cannot be gotten again. I will run 7 separate macro-events. I was curious of the probability that I will get at least 1, 2, 3, etc. desired results.

For example, if the first macro-event is M, and I get a desired result for M₂, it will not be a desired result for subsequent runs, leaving only M₁, M₃, N₁, etc. for the remaining 6 macro-events.

Can somebody help me out on how to solve a problem of this nature?
Thanks

 

[chiro]

I'm trying to figure out the probability of getting something tomorrow, but since I've never taken a statistics course, I don't really know how to do a problem of this nature.

There are two events, I will call the macro-events, M and N for argument sake, each with a 50% chance of happening. In each macro-event, there are also three micro-events, I will call M₁, M₂, M₃, N₁, etc. Each micro-event has a 20% chance of a desired result. After a desired result is reached, it cannot be gotten again. I will run 7 separate macro-events. I was curious of the probability that I will get at least 1, 2, 3, etc. desired results.

For example, if the first macro-event is M, and I get a desired result for M₂, it will not be a desired result for subsequent runs, leaving only M₁, M₃, N₁, etc. for the remaining 6 macro-events.

Can somebody help me out on how to solve a problem of this nature?
Thanks

Hey Nex Vortex and welcome to the forums.

The most intuitive way to get into probability is to use tree diagrams. A tree diagram has branches that go from left to right and events on the right are based on related events that they "inherit" from on the left.

So I'll start with an example of a coin toss that is done twice.

Your first branch will be two main branches: one a head and one a tail. For the head and the tail it will have two branches both a head and a tail. Counting up all the leaves (things with no more branches) you have four events.

The way to calculate a probability for some event (leaf) is to multiply the probabilities at each branch point. Let's say we have a fair coin with equal chance of getting a head or a tail. The probability of getting two heads is 1/2 x 1/2 = 1/4.

Now we could say that because of independence that its easier to use binomial distribution with 2 trials and p = 1/2, but that's not the point of the tree diagram. The point of the tree diagram is to show you how to think about any probability independent or not.

If you are doing events based on time of occurrence (like the coin toss) go from left to right. You should note that at any level of the branch, all leaves in that level have to equal one: the reason is that if they don't then if it is less than one, you have missed out some possible event and if it is greater than one then you have made a mistake.

So start off with your first possible events as your first branches, make sure they add up to 1. Then do your sub-events and make sure all leaves at that level add to one, and repeat until you've gone to whatever level.

Once you've got your tree diagram multiply probabilities at each node right up till the leaf and that is your probability for that event. Note that if you get some events that are dependent on other events happening, this will be reflected in your tree diagram.

So to recap:

1) Draw a tree diagram with each node of the tree being some atomic event (can't divide it any further)
2) Assign a probability to each node
3) Make sure total of all probabilities at a specific tree depth add up to 1
4) For a given event at a particular leaf node multiply all node values from leaf all the way up to trunk to get final probability

and you're done