63% of people are happy. Pick 10 people randomly.
a) What are the odds that 8 of them are happy?
b) What are the odds that at least 2 of them are happy?

Is this a homework question? What have you tried so far?

 

This is a question related to the Bernoulli and Binomial distributions, so you may want to check those out if you haven't already. You could also look at probability trees to give you some intuition, though actually solving problems with them gets messy. There is a simple formula to solve this, but just using it doesn't help understanding.

 

I'm assuming you are sampling from an infinite population - it's different if sampling from a finite population. Do you know which - do you understand the difference?

hey..you can also check out https://www.khanacademy.org/math/probability/probability-geometry/probability-basics/v/basic-probability

63% of people are happy. Pick 10 people randomly.

a) What are the odds that 8 of them are happy?

b) What are the odds that at least 2 of them are happy?

 

Have you ever taken a class in probability? If you did then you should have learned about the "binomial probability distribution". It is called "binomial" because there are two possible outcomes. Often those are called "success" and "failure"- here they are "happy" and "unhappy". If the probability of "success" is p then the probability of "failure" is 1- p and the probability of k "successes" and n- k "failures" in n "trials" is [math]\frac{n!}{k!(n-k)!}p^k (1- p)^{n-k}[/math].

 

"At least 2 happy" is the same as "not only 0 or 1 happy" so the probability that "at least 2 are happy" is 1 minus (the probability none are happy times the probability that 1 is happy).

Edited December 14, 20169 yr by Country Boy

Hello

 

will the solution be different if we choose the people all together or 1 by 1? I think it wouldnt

As long as the independence assumption still holds. Maybe some people aren't happy if picked together.