Question

Two balanced coins are tossed. What are the expected value and the variance of the number of heads observed?
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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.

Answer 1

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

Answer 2

Wait a minute, they said two coins. Assume each tossed once. n = 2

Answer 3

thanks damon for your help