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I've seen many lecture videos that use the example

An experiment consists of 3 coin tosses. The probability of getting a head in any of the tosses is "p"

to explain independence. They say that the coin tosses are independent of each other, but what does that mean? I thought independence was defined on events. The coin tosses are not events. An event is a collection of sample points, and in this example, each sample point consists of a string of 3(the 3 results of each coin toss). So the results of each coin toss are not events per se. Where did I get it wrong?

Also, when we ask "what is the probability that the first toss resulted in heads?" Isn't that the sum of the probabilities of all sample points that have heads as the result of the first toss? But we were told the probability of heads in any toss is "p". How're we sure the sum of the probabilities of all those sample points gives p?

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    $\begingroup$ Three rolls of a die? Then how can the probability of heads be greater than zero, since the set of possible outcomes of each roll is $\{1, 2, \ldots, 6\}$? $\endgroup$ Commented Oct 11, 2023 at 11:05
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    $\begingroup$ Independence can be defined for events (where it's a relatively simple mathematical definition) or for random variables (where it's a slightly more complicated mathematical concept) ; also, we have an intuitive understanding that coin tosses (or die rolls) are "physically independent" in the real world, and therefore if $E$ if an event that depends only on a coin toss, and $F$ is an event that depends only on another coin toss, then if the mathematical definitions are good, it must follow that $E$ and $F$ are independent events, although no math teacher will prove this "theorem" directly. $\endgroup$
    – Stef
    Commented Oct 11, 2023 at 11:11
  • $\begingroup$ Alternatively: if $X$ and $Y$ are two independent random variables, and $E$ is an event defined using only $X$, and $F$ is an event defined using only $Y$, then $E$ and $F$ are independent event. Your math teacher is never going to prove this in an introduction probability class, because that requires defining "independent random variables" and also "event defined using only...", which are a bit too hard formal definitions for an introduction class. $\endgroup$
    – Stef
    Commented Oct 11, 2023 at 11:18
  • $\begingroup$ As for your last question: if you know that the two coin tosses are independent, then the definition of independence tells us that P(first = H) = P(first = H | second = H) = P(first = H | second = T); the law of total probabilities says that P(first = H) = P(first = H | second = H) P(second = H) + P(first = H | second = T) P(second = T) and when putting the two together, along with P(second = T) = 1 - P(second = H), we get P(first = H) = p p + p (1 - p) = p (p + (1-p)) = p. $\endgroup$
    – Stef
    Commented Oct 11, 2023 at 11:28

2 Answers 2

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Two events $A, B$ are independent if $P(A\cap B)=P(A)P(B)$.

In this case you are correct that an event in this experiment is a string of $3$ results. However, when talking about the independence here, we refer to an experiment of a single toss. In this experiment the sample space is $\Omega=\{H,T \}$, and independence refers to the fact that if $A,B$ are subsets of $\Omega$, then $P(A\cap B)=P(A)P(B)$. It is important to mention that we choose the probability $P({H})=p$ (we must specify both the sample space and a probability function to construct an experiment, or, formally, a probability space).

This independence now allows us to construct the probability function of the triple toss experiment, which has the sample space $$\Omega_3=\{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$$

The independence of events in the first exleriment makes it "natural" to define the probability function in the latter experiment in such a way that

$$P_3(\{HHH\})=p^3, P_3(\{HHT\})=p^2(1-p), \space \text{etc.}$$

But, this does not follow directly from the first probability function, it just makes sense to look at $P_3(\{HHH\})$ as equal to $P(\{H\}\cap \{H\} \cap \{H\})$, for example.

As for you last question, checking any probability in the second experiment is done using $P_3$. So the probability that the first roll is $H$ is given by $P_3(\{HHH, HTT, HTH, HHT\})$, which indeed ends up being $p$.

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They say that the coin tosses are independent of each other, but I thought independence was defined on events. The coin tosses are not events. An event is a collection of sample points. So the results of each coin toss are not events per se.

Each coin toss is an experiment trial, and two trials are said to be independent when knowing the result of one does not change the other's results' conditional probabilities. (Note that the trials may be concurrent instead of successive.)

For example, by comparing their columns, can you see that each of these two experiments have dependent trials?

enter image description here enter image description here

But we are really still talking about independence of events: the trial result $H_2$ is independent of the trial result $T_3$ because $P(H_2\cap T_3)=P(H_2)P(T_3)$ because $P(hht,tht)=P(hhh,hht,thh,tht)P(hht,htt,tht,ttt).$

Also, when we ask "what is the probability that the first toss resulted in heads?" Isn't that the sum of the probabilities of all sample points that have heads as the result of the first toss? But we were told the probability of heads in any toss is "p". How're we sure the sum of the probabilities of all those sample points gives p?

Yes, and you are being told that this experiment's trials are independent of one another.

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