Posted by **amelie** on Saturday, February 23, 2008 at 9:52pm.

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.

- Intro to probability -
**Damon**, Saturday, February 23, 2008 at 10:09pm
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

- Intro to probability -
**Damon**, Saturday, February 23, 2008 at 10:10pm
Wait a minute, they said two coins. Assume each tossed once. n = 2

- Intro to probability -
**amelie**, Saturday, February 23, 2008 at 10:39pm
thanks damon for your help

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