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Have you started solving the problem?

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How do you do that?

it is about discrete type random variable.

Yes, so how do you do that with this probability distribution?

\[ Var[unbroken] = P[X=0]*(0-E[X=0])^2 + P[X=1]*(1-E[X=1]) ...\]

Can you place some values in that equation? I'm not sure what E[X=0] stands for?

But what does E|X| stand for?

Expected value of the random variable.

Yes, they are all the expectation. My first formula doesn't really make sense.

So just 5.7? All of them?

Yes, you calculate how far they are from the average.

Ok. Thankyou! :)

E(whole)= 5.7
var(whole)= 0.51

Did I lose you?

0.8 probability of 0 eggs broken means 0.8 probability of 6 whole eggs....

It says the variance is 1.7? Yes, I'm lost.

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