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I have been learning about independent events in statistics.

The textbook we have been given (pg. 12) states that "independent events do not add up to 1".

I'm really confused by this. When I learnt about tree diagrams, all the probabiliteis added to one, like in this example:Independent Events

So why do independent events add to one in that video, while the textbook states that they do not?

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  • $\begingroup$ I am confused too, what is an addition of event ? The addition of their probability ? my guess about this is that you are confusing independence and disjointedness of events, those are very different phenomenon and actually mutually exclusive in typical cases. $\endgroup$ – P. Quinton Feb 17 '19 at 18:07
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    $\begingroup$ You are confusing "independent" with "disjoint". In your picture the two "pick B" events are independent, and the sum of their probabilities is $7/10+7/10=14/10>1$, which should tip you off. $\endgroup$ – kimchi lover Feb 17 '19 at 18:12
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Independent events do not necessarily have probabilities that sum to 1. All that is necessary for two events to be independent are $P(A \cap B) = P(A) P(B)$.

Perhaps you are thinking about a partition. We call a collection of events $A_1, \ldots, A_n$ a partition if $A_i \cap A_j = \emptyset$ for any $1 \leq i, j \leq n$ (that is, all the sets are disjoint) and $A_1 \cup \ldots \cup A_n = \Omega$, where $\Omega$ is the sample space. (In which case their probabilities do sum to 1.)

In my experience students confuse disjointedness and independence; that is, if they try to draw a Venn diagram representing independent events, they draw disjoint events instead. But disjoint events are not independent, because if $A$ and $B$ are disjoint and $A$ occurs, you automatically know that $B$ did not occur.

There's not much that you can say about events who's probabilities sum to one. They need not be independent, and they may not even be a partition. Probability allows for sets that are not empty but still have probability zero.

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Two events $A$ and $B$ are defined to be independent if one of them happening doesn't interfere with the other. Mathematically this is expressed by

$$ P(A \cap B) = P(A)P(B) $$

To illustrate, let's say $A$ = It will rain tomorrow and $B$ = You pick a card from a deck and it's a King. It's obvious these events are independent and, clearly, you can have $P(A) + P(B) > 1$. So, no, independent events do not necessarily add up to $1$, but it may happen by coincidence.

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  • $\begingroup$ Thanks for your answer. I read this link mathsisfun.com/data/probability-tree-diagrams.html and it said that the probability of all the events add to 1? $\endgroup$ – Christopher U Feb 17 '19 at 18:36
  • $\begingroup$ If you flip a coin two times, the Sample Space that is generated is exactly $HH, HT, TH, TT$. The probability of all events of a sample space must always sum up to $1$. $\endgroup$ – Victor Feb 17 '19 at 18:42
  • $\begingroup$ So with those tree diagrams, do the probabilities always add to 1? I'm just confused about when they add to one and when they don't. This website says they always add to one www-math.bgsu.edu/~albert/m115/probability/prob_rules.html $\endgroup$ – Christopher U Feb 17 '19 at 19:09
  • $\begingroup$ It is not incorrect. Remember my P(A) + P(B) example. $\endgroup$ – Victor Feb 17 '19 at 19:17
  • $\begingroup$ Coin flips are independent events because if you flip a coin, then flip it again, the result of the first flip does not interfere in the result of the second flip $\endgroup$ – Victor Feb 17 '19 at 19:17

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