1. ## Probability word problem

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?

2. ## Re: Probability word problem

this is a binomial distribution with parameters $p=0.63,~n=10$

can you solve it now?

3. ## Re: Probability word problem

Where did you get this problem? It is strange that you should be given such a problem without having learned anything about problems like this. As romsek said, this is a pretty standard "binomial distribution" problem. Do you know what that is? Even if you do not, it is not too difficult to 'reason it out'. The probability that the first 8 people chosen are happy is $\displaystyle 0.63^8$. The probability the other two are not is $\displaystyle 0.37^2$. The probability the first 8 are happy and the last two are not is $\displaystyle (0.63^8)(0.37^2)$. In fact, it is not difficult to see that the probability that 8 are happy and 2 are not in any specific order is also $\displaystyle (0.63^8)(0.37^2)$. But there are 10!/8!2!= 45 different orders.

To find the probability that "at least 2" are happy, find the probability that none are happy and that 1 is happy, add to find the probability that either none or 1 is happy then subtract from 1.

4. ## Re: Probability word problem

Originally Posted by HallsofIvy
$\displaystyle (0.63^8)(0.37^2)$
Are those exponents?

$\displaystyle (0.63^8)$ = 0.02481557802?

$\displaystyle (0.37^2)$ = 0.1369?

5. ## Re: Probability word problem

Originally Posted by romsek
this is a binomial distribution with parameters $p=0.63,~n=10$

can you solve it now?
0.1528764?

6. ## Re: Probability word problem

Originally Posted by MathLearner2016
Are those exponents?

$\displaystyle (0.63^8)$ = 0.02481557802?

$\displaystyle (0.37^2)$ = 0.1369?
Yes, those are correct. Now multiply them together to get the probability of any specific ordering of 8 and 2 things, then multiply by 45 to get the probability of all orders.

7. ## Re: Probability word problem

Originally Posted by MathLearner2016
0.1528764?
Yes, to 7 decimal places $\displaystyle 45(.63)^{8}(.37)^2= 0.1528764$. Why that specific accuracy? Since you are given the probability to only two decimal places, I would have written 0.15.

8. ## Re: Probability word problem

b)

So the probability of nobody being happy is 0.00005; one being happy is 0.0008 and two being happy is 0.0063. Can I use that to figure out the probability of at least 2 being happy?