Intro to probability
posted by amelie .
Two balanced coins are tossed. What are the expected value and the variance of the number of heads observed?
I'm confused on how to start. I don't know how many times im suppose to toss coin, and how to set up my table for the expected value and then find variance, I know how to find expected value when I know how to use what's given to me, hope you can help. I feel that the answer is probably simple but i'm stuck on info i think it lacks.
p = 1/2 for heads or tails
this is a binomial distribution
mean = n p = n/2
variance = n p (1-p) = n/4
It is indeed lacking n so has to be answered that way
Wait a minute, they said two coins. Assume each tossed once. n = 2
thanks damon for your help