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Repeated Trial question with bernoulli equations

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Full Member level 4
Aug 19, 2005
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repeated trial

A box contains 1000 bulbs of which 100 is defective, two bulbs are picked at random, find the probability that both are defective.

If we perform :

The probability that the first is defective is P(1D) = 100/1000 = 0.1
The probability that the second is defective is P(2D) = 99/999 = 0.099099...

thus the probability that both are defective is P(1D)*P(2D) = (100/1000)*(99/999). I think this is the correct result.

Now if we perform by bernoulli equations:

We have n = numbers of bulbs = 1000
We have m = number of bulbs taken = 2
We have p = Probability that the bulbs are defective equals = 100/1000=0.1

Applying Bernoulli:

P(2D) = [C(n,m)]*[p^m]*[(1-p)^(n-m)] = (1000*999/2)*[(0.1)^2]*[0.9^998] this result is cleary wrong. So here comes the question, why ? what is wrong ?

Re: Repeated Trial question

There is a confusion of notions. That is about P(2D). I suppose the P(2D) in "P(2D) = 99/999 = 0.099099..." means the probability about the second, while P(2D) in "P(2D) = [C(n,m)]*[p^m]*[(1-p)^(n-m)]..." means the probability of both the first and the second.

Besides, when you say "Applying Bernoulli:", I suppose you want to say "Applying Binomial:". Binnomial is the successive tests of INDEPENDENT Bernoulli's. Notice that I have capitalized the word - "independent". That is, if you want to apply "binomial", the successive test must be independent. Apparently, your teats are not indepenedent, Therefore, this is wrong:

claudiocamera said:

Applying Bernoulli:

P(2D) = [C(n,m)]*[p^m]*[(1-p)^(n-m)] = (1000*999/2)*[(0.1)^2]*[0.9^998] ...
Repeated Trial question

Thanks for the answer, cleary the problems was the dependence of the events , due to no replacement of the bulb picked.
Unfortunatelly I came across another doubt: If we consider that we picked the first bulb, returned it to the box and repeated the experiment and then asked what is the probability that in both trials the bulbs picked are defectives. We can apply bernoulli, since we have replacement and independent events, so performing the exercise by two methods:

The probability that in each trial the bulb is defective is P(D) = 100/1000 = 0.1 thus The probability that in both trial we pickec defective bulbs is P(D)*P(D) = 0.1 *0.1 = 0.01.

Now applying bernoulli:
P(D) = [C(n,m)]*[p^m]*[(1-p)^(n-m)] = (1000*999/2)*[(0.1)^2]*[0.9^998].

Again the result is completely different !? What is wrong?

Re: Repeated Trial question

Here is what happened.

First of all, what are the two numbers, 100 and 1000? These two numbers play only one role, that is, they serve to determine the probabilities of bad (p=0.1) and good (q=0.9) bulbs in the box. None of them has anything to do with the number of bernoulli tests. That is, you are not performing 1000 times of bernoulli tests.

Actually, you are performing only two consecutive bernoulli tests, because those are the only number of bulbs you picked. Therefore, n=2. And you are asking the probability of all bad bulbs in the two tests. Therefore, m=2. Thus, the probability by applying successive bernoulli tests is


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