Bernoulli and binomial distribution

[ps_ricky_son]

hi all

can anyone help me to distinguish the 2 types of prob. distribution?

Thx a lot

 

[David Harper CFA FRM]

Hi @ps_ricky_son I don't know if you attempted a search first, but a binomial distribution is a series of independent Bernoullis, see http://en.wikipedia.org/wiki/Binomial_distribution
The classic risk application is a portfolio of credits (where defaults are characterized by binomial) under the unrealistic assumption each credit is i.i.d. binomial. Thanks,

 

[ps_ricky_son]

thx David~~~ that means when we are talking the whole model, we are referring to binomial distribution. But, if we are talking about each variable inside that model, the case will be independent Bernoullis? thx again

 

[David Harper CFA FRM]

Hi @ps_ricky_son I myself wouldn't use that language, I think the wikipedia entry is precise. Please note the binomial distribution entry reads, "In probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent yes/no experiments, each of which yields success with probability p. A success/failure experiment is also called a Bernoulli experiment or Bernoulli trial; when n = 1, the binomial distribution is a Bernoulli distribution. The binomial distribution is the basis for the popular binomial test of statistical significance."

So another way to think of this is: the Bernoulli is a special case of a binomial with n = 1. Thanks,

 

[ps_ricky_son]

that is more clear to me ~~ thanks David

 

[mixelfg]

Hi,

Not sure how to start a new thread so seeing as it is related can you please help with the following question:

Q:
A multiple choice question has 10 questions, with five choices per question. If you need at least three correct answers to pass the exam, what is the probability that you will pass simply by guessing?
a. 0.8%
b. 20.1%
c. 67.8%
d. 32.2%

Answer being d.

My calculation was = 10c3*(0.2^3)*(0.8^7) = 20.1% - so i choose b....

What am I doing wrong? how is the correct answer derived please?

Thanks!

 

[ShaktiRathore]

P(c)=prob of correct answer picking out of 5=1/5
P(p)=prob to pass exam=1-prob of failing
prob of failing is choosing 0or1or2 correct answers out of 5=10C0*.2^0*.8^10+10C1*.2^1*.8^9+10C2*.2^2*.8^8
So P(p)=1-( 10C0*.2^0*.8^10+10C1*.2^1*.8^9+10C2*.2^2*.8^8)
Thanks

 

[mixelfg]

Thank you so much for the explanation! Makes sense.... I went about it all wrong.. More clear now.