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 Post subject: Word problem with probabilities and averages
PostPosted: Tue, 15 Nov 2011 09:01:50 UTC 
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I have four scenarios with five rounds each where an event x, y, or z will be chosen.

In the first scenario, each round has a equal chance of triggering event x and will continue triggering event y each round until event x is triggered. Example1: if event x is triggered on r1, y=0. Example2: if event x is triggered on r3, y=2. Example3: if event x is triggered on r5, y=4.

In the second scenario, each round has a 20% chance of triggering event x and will continue triggering event y each round until event x is triggered. Example: event x is triggered on r1, y=0. Example2: event x is triggered on r4, y=3. Example3: event x is never triggered, y=5.

In the third scenario, each round has a 75% chance of triggering event x and if it is not triggered that round it will trigger event y. Example1: event x is triggered every round, y=0. Example2: event x is triggered on rounds 1,2,and 5, y=2. Example3: event x is triggered on rounds 3 and 4, y=3.

In the fourth scenario, rounds 1 through 4 each have a equal chance of triggering event x. Each round x has not been triggered, there is a 50% chance that event y will trigger else event z will trigger. Once x has been triggered, event y will not trigger. Example1: event x triggers on round 1, y=0. Example2: event x triggers on round 4, y triggers on round 1 and 3, y=2.

I am trying to learn how to calculate the average on the amount of times event y will trigger in each scenario based on the probabilities given in each scenario. I took calculus in high school so I am a little rusty on my math, but if someone explained it I would understand.

Here are my attempts:

Scenario #1
If each round has an equal chance of triggering event x, and there are five rounds, that would mean that each has a 20% chance to trigger. So I would think the formula would be:
((.2*0)+(.2*1)+(.2*2)+(.2*3)+(.2*4))/5 = 40% chance to trigger y each round

Scenario #2
I am not sure how to do this one as it should be different from scenario 1, and I think I messed up on scenario 1.

Thank you for any help you can give me.


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 Post subject: Re: Word problem with probabilities and averages
PostPosted: Wed, 7 Dec 2011 18:42:49 UTC 
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Joined: Tue, 20 Nov 2007 04:36:12 UTC
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Location: Las Cruces
lisaburlew wrote:
I have four scenarios with five rounds each where an event x, y, or z will be chosen.

In the first scenario, each round has a equal chance of triggering event x and will continue triggering event y each round until event x is triggered. Example1: if event x is triggered on r1, y=0. Example2: if event x is triggered on r3, y=2. Example3: if event x is triggered on r5, y=4.


Your description of the problem is unclear. If y is an event, what kind of event is it? Is it something that either happens or doesn't? If that is the case, why do you give an example with y = 4? Are you using "y" to denote both the event and the total number of times that it happened? (If so, stop doing that!) Are x,y and z mutually exclusive events? Or is it the case that several of these events can happen on the same round?

Quote:

I am trying to learn how to calculate the average on the amount of times event y will trigger in each scenario



It isn't clear what you count as a "time". Does event y happen or not happen on each round? If it happens on rounds 2,4 and 5, do you count that as three times that it happened? -or do you count that as one time since it did happen at least once in the scenario?


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