ChatterBank0 min ago
Is It A Binomial Experiment ?
4 Answers
There is an example :
A box contains 20 cell phones, and two of them are
defective. Three cell phones are randomly selected from this
box and inspected to determine whether each of them is good
or defective. Is this experiment a binomial experiment?
if I know that:
A binomial experiment must satisfy the following four
conditions:
1. There are n identical trials.
2. Each trail has only two possible outcomes.
3. The probabilities of the two outcomes remain constant.
4. The trials are independent.
My question is :
1- what do the last two points ( conditions ) mean ?
2- why we consider the example above not to be a binomial experiment ?
A box contains 20 cell phones, and two of them are
defective. Three cell phones are randomly selected from this
box and inspected to determine whether each of them is good
or defective. Is this experiment a binomial experiment?
if I know that:
A binomial experiment must satisfy the following four
conditions:
1. There are n identical trials.
2. Each trail has only two possible outcomes.
3. The probabilities of the two outcomes remain constant.
4. The trials are independent.
My question is :
1- what do the last two points ( conditions ) mean ?
2- why we consider the example above not to be a binomial experiment ?
Answers
Best Answer
No best answer has yet been selected by cilineelzein02. Once a best answer has been selected, it will be shown here.
For more on marking an answer as the "Best Answer", please visit our FAQ.It would only be a binomial experiment if, for some unaccountable reason, the first phone to be examined was replaced in the box before a second selection was made (and likewise with the second phone). Under those circumstances P(D), where D is a defective phone being found, would remain constant at 3/20.
However (although it's not explicitly stated), the assumption here must be that a phone isn't returned to the box after inspection. If so, then when the second phone is taken, P(D) will be either 3/19 or 2/19, depending upon the condition of the first phone. i.e. the probabilities P(D) and P(D') don't remain constant because each trial is dependent on the outcome of any preceding trial(s).
However (although it's not explicitly stated), the assumption here must be that a phone isn't returned to the box after inspection. If so, then when the second phone is taken, P(D) will be either 3/19 or 2/19, depending upon the condition of the first phone. i.e. the probabilities P(D) and P(D') don't remain constant because each trial is dependent on the outcome of any preceding trial(s).
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