Q:random variable that follows binomial distribution

Schweser Questionbank ID- #1479 Which of the following could be the set of all possible outcomes for a random variable that follows a binomial distribution? A) (-1, 0, 1). B) (0, 0.5, 1, 1.5, 2, 2.5, 3). C) (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11). D) (1, 2). The correct answer was C) (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11). This reflects a basic property of binomial outcomes. They take on whole number values that must start at zero up to the upper limit n. The upper limit in this case is 11. This question tested from Session 3, Reading 9, LOS d, (Part 2) anyone can explain this?

I am too not able to understand this. I remember doing it wrong too.

The binomial distribution is a probablillity distribution of number of successes for a given number of trials. Each trial can either result in failure (value=0) or success (value =1). You can think of it as a coin toss, each time you toss the coin you can either win (1) or loss (0). If you were to list possible outcomes it would then always be greater then 0. (you can’t win -1 coin tosses!) And the number have to descrete, ie you can’t win 1.5 coin tosses. The distribution must always start at zero. This makes sense as if you were listing all the possible outcome that could happen. The least number of wins will always be zero. C is the only answer that does not brake any of these “rules”

I see, it is asking all possible outcomes for a binomial distribution random variable . like toss coin 11 times, you may get 0 time tail, or 1, or 2…or 11 times. tks!

Yes. This question is horribly worded. Keyword is distribution.

thw q’s wording is horrible.

but the logic makes sense. thanks dude.