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I struggle defining the experiment, the sample space and the events.

For instance, I have two ways of defining the following:

  1. I have the experiment of tossing a coin twice, my sample space would be {HH, HT, TH, TT}. If I have two events, A: Getting H in the first toss, B:getting H in the second toss. Then A={HH, HT} and B={HH, TH}, where P(A)P(B) = 1/4.

  2. I have the experiment of tossing a coin that is repeated twice, the sample space is {H,T}. I have two events, A:Getting H in the first toss, B:getting H in the second toss. Then A={H} and B={H}. where where P(A)P(B) = 1/4. Although I find this weird, it is how I see commonly calculated probabilities, but the experiment and events are not usually explicitly defined. (Please correct me if I am wrong).

If I want to give an example with two dependent events, then I could say the following:

  1. I have the experiment of tossing a coin twice, my sample space would be {HH, HT, TH, TT}. If I have two events, A: Getting H in the first toss, B:I don't know how to define this event.

I don't want to call B:getting H in the second toss given that I got H in the first toss, because that is represented with P(B|A) and I need to specify what B is.

I also don't want to call an event like Getting two Hs because I can't identify A and B.

So, how can describe the events? and the experiment? and the sample space?

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1 Answer 1

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In point 2), _ the experiment of tossing a coin that is repeated twice_, this is the same as 1) experiment of tossing a coin twice. The sample space is {HH, HT, TH, TT}. Note: there are four possible events, which are the individual elements of your sample space.

In your "two dependent events" example, there are also four possible events, which are the same set as in 1).

What differs between independent and dependent scenarios is the conditional probabilities. If $B$ is the event of getting H on the second toss, then:

  • In an independent tosses scenario, $P(B) = \sum_n P(B ,A_n ) = \sum_n P(B|A_n)P(A_n)$ (law of total probability, where $A_n$ enumerates over the different outcomes of the first coin toss, H or T). $P(B|A_n) = P(B)$ due to independence, so $P(B) = \sum_n P(B)P(A_n) (0.5 \times 0.5) + (0.5 \times 0.5) = 0.5$ (assuming a fair coin)
  • In a dependent tosses scenario, still $P(B) = \sum_n P(B|A_n)P(A_n)$, but we you have to know further information about $P(B|A_n)$ to be able to solve for $P(B)$. For example, see the "tree diagram" plot at this link.
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