# Basics of probability - Independent Events.

I read that the probability of two events provided they are independent is obtained by the following formula:

$Probability _ {Independent~ Events} = Probability_{1st Event} \times Probability_{2nd Event} \times ...$

Now it states that the probability of getting a Tails and Head (T-H) or getting a Heads and Tails (H-T) when a coin is flipped consecutively is given by:

$P(H-T) + P(T-H) = (\frac{1}{2}\times \frac{1}{2}) + (\frac{1}{2}\times \frac{1}{2}) = \frac{1}{2}$

Now my question is why isn't it

$P(H-T) \times P(T-H)$ instead of $P(H-T) + P(T-H)$ since after all the two events are independent of each other. I would appreciate it if someone could clear this up.

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These two events actually highly dependent. Do you see why? –  Artem Aug 5 '12 at 23:14
Are you talking about P(H-T) and P(T-H) ? –  MistyD Aug 5 '12 at 23:15
I am talking about events H-T and T-H. –  Artem Aug 5 '12 at 23:16
How are are they dependent ? –  MistyD Aug 5 '12 at 23:17
Events are only dependent if an event that occurs has a probability that is different solely based on a preceding event. For example, if you pick a card out of a deck, and then a second card out of the same deck without putting the first card back, that would be a dependent event. –  mathguy Aug 5 '12 at 23:37
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Well..Independence and dependence has nothing to do with the fact that you are adding up the probabilities. You are asking what the probability that either THIS happens OR THAT happens. When you see the word "or" that means you are supposed to add up the probabilities (given that they do not overlap).

Now the statement you made is true, and it is partially applying to your question (particularly to each individual event). Your statement applies to situations like "what is the probability that THAT will happen AND THIS will happen. Let's take the first set of events: H-T. Whatever side you get on the second flip is completely independent of the first flip. The chance of getting heads on the first flip is 1/2, and the change on getting tails on the second flip is 1/2, so you multiply them together to get 1/4.

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One way to help visualize this problem might be to list out all of the possible outcomes of flipping a coin twice. There is a much more rigorous statement to be made which is hinted at above about the independence and dependence, but think of it this way.

If you are trying to assess $P(Heads, then\;Tails)$ or the reverse, think about this:

Your probability space $\Omega$ consists of all possible outcomes of 2 consecutive coin tosses. Since there are only 4 possible outcomes in this example, we list them explicitly:

$\Omega=\{x,y\}:x\in H,T\;\;y\in H,T$

$\Omega= \{\{H,T\},\{H,H\},\{T,H\},\{T,T\}\}$

The above lists each possible outcome, and the associated probability for each is $\frac14$. Now if you want to know the combined probability that you flip Heads, then Tails -OR- Tails then Heads, it is simply the addition of each probability.

But if you wanted to know the probability of flipping heads, and then flipping tails, since with a fair coin the events are independent, we just apply the formula you mentioned and multiply each event's probability together $\frac12 \times\frac12=\frac14$

The reason you might be getting confused on this is because the odds of flipping Heads then Tails is highly dependent upon whether or not you flip heads on the first coin.

Does that make sense?

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Yes it does. However I am currently sticking to the fact that if I need an "or" then I'll add otherwise I''ll multiply –  MistyD Aug 6 '12 at 1:20
nothing wrong with that for basic problems, but in case things get a little more complicated hopefully this is of some value :) GOod luck –  Justin Aug 6 '12 at 1:38
You use the term $\it{or}$, so the question you are trying to answer has nothing to do with whether the two events are independent or not. The formula you gave means that the probability that two events occur $\it{simultaneouly}$ can be written as the product of the probabilities of the one event and of the other event if these events are independent.