# Binomial Distribution Question

Here’s the Question: Which of the following could be the set of all possible outcomes for a random variable that follows a binomial distribution? A) (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11). B) (-1, 0, 1). C) (0, 0.5, 1, 1.5, 2, 2.5, 3). D) (1, 2). My answer was D, because there are only two possible outcomes for a binomial distribution. But the correct answer given was A, saying that binomial outcomes take on whole number values that must start at zero up to the upper limit, n (in this case 11). Any help explaining this would be great. I might just be a moron, but a little help would be appreaciated.

it is simple. You are right when you says that there are only two possible outcomes for a binomial distribution, but it is also true that these two outcomes can be one of the n possible finite numbers (i.e. from 0 to n) therefore, A is the correct. I hope this help.

scottrader, you are confusing binomial distribution and bernoulli distribution. http://en.wikipedia.org/wiki/Binomial_distribution http://en.wikipedia.org/wiki/Bernoulli_distribution A is the correct answer.

maratikus Wrote: ------------------------------------------------------- > scottrader, > > you are confusing binomial distribution and > bernoulli distribution. > > http://en.wikipedia.org/wiki/Binomial_distribution > > http://en.wikipedia.org/wiki/Bernoulli_distributio > n > > A is the correct answer. Maratikus, Thanks for your explain but sorry for trouble you with my confusion. I still don’t understand much why should A is the correct answer but C. I think C may be the correct answer as your explain. The reason why A should be the best choice because the number of possible outcomes of A is 12 and this number 12 can be balance with failure and success outcomes in binomial distribution. I mean there will be 6 failure outcomes balance with 6 success outcomes in trials and it will satisfy the binomial distribution condition. The other three answer cannot be outcome of binomial distribution because they are odd number of outcomes. Am i right or wrong understanding? Thanks for your comments

Thu Thuy, The outcomes of binomial distribution represent the number of successes and, therefore, are non-negative integers. B is out because -1 is a negative number C is out because there are non-integers there D is out because 0 should be there (because set of ALL possible outcomes). A is the answer. I hope that helps.

scotstrader, just curious where did u find this question?

maratikus Wrote: ------------------------------------------------------- > Thu Thuy, > > The outcomes of binomial distribution represent > the number of successes and, therefore, are > non-negative integers. > > B is out because -1 is a negative number > C is out because there are non-integers there > D is out because 0 should be there (because set of > ALL possible outcomes). > > A is the answer. > > I hope that helps. Maratikus, Pls kindly correct me if i am going wrong. If instead of marking number from 1 to 6 to the six sizes of a die, i will mark the number from -3, -2, -1, 1, 2, 3. Then, i think the group (-3, -2, -1, 1, 2, 3) are the possible outcomes of rolling a die. And if i called getting numbers (-3, -2, -1) are failure and (1,2,3) are success of rolling a die? I can say the value of the outcomes are not effected anyway. And the above outcomes for each case is the same. As i think of it by this way, so i still think the balance of number success and failure trial will be determine the binomial distribution. You mean your explain above the value of outcome itself, right? Could you pls make it more clearly? Thanks a lot for your explain. I get so confused this part of quantitative.

Thu, Output of the binomial distribution is the number of occurences, not the name/value of the actual events.

The marking of the die is irrelevant - three sides of the die are successes and three sides are failures (could be 4/2, 5/1, even 6/0). If you roll the die 10 times and count the number of successes you have a binomial distribution. With a “fair” die, p(success) = 1/2 in this case. But the possible outcomes of this experiment are 0, 1, 2, …, 10. Now if we did something else like summed the numbers that show up, we would have some different kind of random variable.

This is another terribly worded question. Are there questions like this on the exam?

I don’t think it’s terribly worded (in fact, it’s mathematically very precisely worded unlike most of Schweser stats questions). In fact, I think it’s a really standard question from introductory stats books everywhere. So, yes, there will be questions like this on the exam.