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There is one error in one of five blocks of a program. To find the error, we test three randomly selected
blocks. Let $𝑋$ be the number of errors in these three blocks. Compute $E(X)$ and $Var(X)$.

That's what i've tried:
$p = 0.6$
$n = 1$
$E(X) = ncdot p = 1cdot (0.6) = 0.6$
$Var(X) = n cdot pcdot q = 1cdot (0.6)cdot (0.4) = 0.24$

If we were using $n = 2$ with success probability of $0.6$ we would have:
$P(X = 0) = 0.16$
$P(X = 1) = 0.48$
$P(X = 2) = 0.36$

$p = 0.6$
$n = 2$
$E(X) = 2cdot (0.6) = 1.2$
$Var(X) = 2cdot (0.6)cdot (0.4) = 0.48$

Since there is one error in the 5 blocks of code, $X in {0,1}$, where $X=0$ if all 3 blocks of code being tested are correct.

Assuming the error is equally likely to be in any of the 5 blocks of code, the probability of drawing three correct blocks of code is $frac{4}{5} times frac{3}{4} times frac{2}{3} = frac{2}{5}$, i.e. $P(X=0) = frac{2}{5}$ and $P(X=1)=frac{3}{5}$.

$mathbb{E}[X] = 0 times frac{2}{5} + 1 times frac{3}{5} = frac{3}{5}$.

$mathbb{E}[X^2] = 0^2 times frac{2}{5} + 1^2 times frac{3}{5} = frac{3}{5}$.

$Var(X) = mathbb{E}[X^2] - mathbb{E}[X]^2 = frac{3}{5}-left( frac{3}{5} right)^2 = frac{6}{25}$.