Which of the following experiments are Bernoulli trials?

a. A die is tossed 4 times and we record the number obtained for each trial.

b. A die is tossed 4 times and we record whether or not the number obtained is odd for each trial.

c. A box contains 4 red and 6 blue balls. A ball is picked up from the box 3 times with replacement.

d. A box contains 4 red and 6 blue balls. A ball is picked up from the box 3 times without replacement.

My Work: I know the following

Bernoulli trials (p: parameter) have the following properties:

a. Each trial yields one of two outcomes: success (S) or failure (F).

b. P(S) = p and P(F) = 1-p

c. Each trial must be independent

So using the above I said

A) Not Bernoulli Trials as although the trials are independent, there's no success or failure defined here. What number do we identify as a success? It's not given so we can't apply Bernoulli Trials here.

B) Yes, this is Bernoulli Trials as there's a definitive success (Getting a odd number) and failure (getting a even number) and the trials are independent as well!

C) No...? There's no success condition or failure condition defined here (What's a success? what's a failure?) so not Bernoulli Trials. Although they are independent trials.

D) No, same reason as C and add on the fact that the trials aren't independent as there's no replacement.

Is my reasoning/logic correct? Especially with C and D? I feel like I may be overthinking it. Any help with better identifying Bernoulli Trials in experiments would be greatly appreciated!

  • $\begingroup$ Indeed, the weird thing is that in C and D, the random variable is not even defined. So I agree with you: A is not, B is, and C and D are poorly defined. $\endgroup$ – Matti P. Aug 10 '20 at 6:10
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    $\begingroup$ I think it's kinda obvious that the random variable in C and D is supposed to be the color of the ball you picked. Then just pick one of the two colors as success and the other as failure. The point of a Bernoulli trial is not really that there is success and failure, but that there are exactly two possible outcomes. $\endgroup$ – Vercassivelaunos Aug 10 '20 at 6:51

I'm going out on a limb here, because I've never heard the phrase "Bernoulli trials". However, I think that I can reverse engineer an answer based on the OP's work and based on Matti P.'s comment, which I partially disagree with.

In part b, it is apparently okay (as a matter of protocol) to arbitrarily construe an odd number as a successful event. Based on this, I would argue that in both parts c and d, it is okay to construe a red ball as a successful event.

Then, (again based on the OP's problem and Matti P's comment), I would distinguish between parts c and d as follows:

In part c, with the sampling done with replacement, the outcome of each sampling is independent of the outcome of any of the other samplings. Based on this, I (arbitrarily, and perhaps wrongly) construe part c as a bernoulli trial.

In part d, the sampling is done without replacement. This means that the outcome of each sampling is not independent of the outcome of the other samplings. Therefore, based on my construance, part d is not a bernoulli trial.

I confess that I metacheated here.
I admit that the distinction between parts c and d could be a red herring, and the problem could represent a carefully constructed trap. However, based on the OP's posting, since a core characteristic of bernoulli trials is independence, my instinct is that my construance above is intended by whoever formulated the question.


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