Caculate the probability using a binomial distribution

[Cyannaca]

Ok so I have a problem I am not sure of the method I should use. In a recent survey, 60% of the population disagreed with a given statement, 20% agreed and 20% were unsure. Find the probability of having at least 5 person who agree in a mini-survey with 10 people.

I tried to caculate the probability using a binomial distribution with n=10, p=0,2 agree and 1-p = 0,8 who either agree or are unsure, and

P(X) = (n!/ (k!(n-k)!)) (p^x) ((1-p) ^(n-x))

I added p(5), p(6)... p(10) and I got p(total) = 0,0328

Is it right to use a binomial distribution in this case?

 

[Dick]

Yes, that is the correct distribution to use.

 

[cepheid]

Yes. The number of people agreeing with the statement in n trials is random variable with a binomial probability distribution. This is because each individual event or trial has two possible outcomes (agreement or not agreement, if you choose to group them that way), and as a result is described by a Bernoulli random variable.

 

[Cyannaca]

Thank you!