I'm learning a statistics tutorial which considers $$ X_i \sim \mathit{Bernoulli}\left(\frac{\theta}{3}\right) $$ to describe an experiment where one draws a ball 4 times with replacement from a bag that contains 3 balls. And then the tutorial gives another experiment where $$ X_i \sim \mathit{Binomial}(3, \theta) $$ and $(x_1,x_2,x_3,x_4)=(1,3,2,2)$.

I understand $X_i \sim \mathit{Binomial}(3, \theta)$ when considering 3 the number of experiments where one experiment/trial flips one coin one time and $\theta$ denotes the success possibility for each trial. but I don't understand the definition of one experiment in that tutorial.

Could someone give me a hint?

Possible setting 1

Does 3 in $X_i \sim \mathit{Binomial}(3, \theta)$ mean there are 3 balls in the bag?

Note that in the second experiment, 3 blue balls are observed which means there already are 3 blue balls in the bag, which means, all 3 balls are blue, if the 3 in $X_i \sim \mathit{Binomial}(3, \theta)$ means there are 3 balls in the bag.

Alternatively, the experiments are using different bags, but that doesn't seem to be compatible with the definition of binomial distribution.

Possible setting 2

Does it make sense to define one experiment as follows?

"draw a ball, record if it's blue, and put it back in the bag, repeat 4 times"

In other words, one experiment draws 4 times, so the possible outcomes are ${0, 1, 2, 3, 4}$. And then, $X_i \sim \mathit{Binomial}(3,\theta)$ means doing the experiment 3 times.

In this setting, $\theta$ doesn't mean the possibility that one blue ball is observed, so what does $\theta$ mean?

Possible setting 3

Another possible definition of one experiment could be

"draw 3 balls from the bag, record how many blue balls observed, and then put the balls back in the bag"

So the possible outcome of one experiment could be ${0, 1, 2, 3}$. See the difference between this setting and setting_1?

The 3 in $X_i \sim \mathit{Binomial}(3, \theta)$ means doing the experiment 3 times.

Again, what does $\theta$ mean in this setting?

  • 2
    $\begingroup$ Example 8.7 is a totally different exercise from the example 8.8. In the $8.7$ it clearly tells you that $\theta$ is the number of blue balls, in order to have a proper probability for the Bernoulli $\theta/3$, whereas in $8.8$ has nothing to do with balls experimenets. $\endgroup$
    – Fiodor1234
    Sep 4, 2021 at 12:00
  • $\begingroup$ @Fiodor1234 Thank you. To help me in understanding the concepts, could you please take the examples of drawing balls to make example 8.8 more concrete? or any other more concrete examples, to illustrate what $(x_1,x_2,x_3,x_4)=(1,3,2,2)$ and the corresponding $\theta$ mean? $\endgroup$
    – soplus2018
    Sep 4, 2021 at 15:19
  • $\begingroup$ I no longer understand your question after your edits, unfortunately. You seem to be fixated on learning the technical definition of the word 'experiment' in this context, but there is none. Don't put too much stock in this tutorial's use of that word. $\endgroup$
    – psboonstra
    Sep 6, 2021 at 1:37

2 Answers 2


Does 3 in $X_i\sim Binomial(3,\theta)$ mean there are 3 balls in the bag?

No. If you want to apply Example 8.8 to the 'balls in the bag' example that the tutorial gives earlier, the 3 means that you do the following action three times: draw a ball, record if its blue, and put it back in the bag (this last part is very important). The total number of blue balls you record is $X_i$ and is a binomial random variable. It will take on values of 0, 1, 2, or 3. You could do this if there were just one ball in the bag or if there were 100 balls in the bag.

In example 8.8, it is telling you that the entire process in the paragraph above was repeated 4 times, and hence you have a set of 4 numbers, each of which could be 0, 1, 2, or 3.

It was perhaps unwise of the tutorial author to use $\theta$ for both the $Bernoulli(\theta/3)$ example (where $\theta$ is the number of blue balls in the bag) and the $Binomial(3,\theta)$ example, where $\theta$ represents the probability of success (i.e. drawing one ball and getting a blue ball).

=====Addition in response to OP's edits=====

A Binomial with $n$ trials is defined as the sum of $n$ independent Bernoulli random variables having a common probability of success, and the sum of independent Binomial random variables each having a common probability of success is itself a Binomial random variable with number of trials equal to the sum of the constituent number of trials. Thus, if $\mu$ is a probability of success (I'm changing the parameter from $\theta$ to $\mu$ so that it is clear I'm not using the tutorial's parametrization), all of the following experiments yield equivalent statistical information (and have the same likelihood up to a multiplicative constant) about $\mu$:

(i) 20 independent draws from $Bernoulli(\mu)\equiv Binomial(1,\mu)$

(ii) 10 independent draws from $Binomial(2,\mu)$

(iii) 5 independent draws from $Binomial(4,\mu)$

(iv) 1 draw from $Binomial(20,\mu)$

The parameter $\mu$ has the same interpretation in all cases: the probability of success for a single Bernoulli random variable.

  • $\begingroup$ Thank you so much. Does it make sense to define one experiment as follow: draw a ball, record if it's blue, and put it back in the bag, repeat 4 times? In other words, one experiment draws 4 times, so the possible outcome is {0, 1, 2, 3, 4}. $X_i\sim Binomial(3,\theta)$ means doing the experiment 3 times, is my understanding correct? $\endgroup$
    – soplus2018
    Sep 2, 2021 at 23:00

The parameters of the binomial distribution are n, the number of independent trials and p, the probability of success on each trial.

So here, the three in the second experiment means that three balls will be drawn from the bag. In this setup, they must be drawn with replacement, because otherwise the success probability would change each time and the binomial distribution would be inappropriate.

I'll have to assume that drawing a blue ball counts as a "success" here. I would say there must be more than three blue balls in the second experiment, because in the first result, x1 is 1. This means at least 2 non-blue balls were drawn.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.