A box contains 8 red and N green balls. Two b | vedclass

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(C) Total number of balls $= 8 + N$.Number of balls drawn $= 2$.The random variable $X$ represents the number of green balls, so $X \in \{0, 1, 2\}$.T

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$12$

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(C) Total number of balls $= 8 + N$.Number of balls drawn $= 2$.The random variable $X$ represents the number of green balls, so $X \in \{0, 1, 2\}$.The probability distribution is given by hypergeometric distribution:$P(X=k) = \frac{\binom{N}{k} \binom{8}{2-k}}{\binom{N+8}{2}}$.For a sample of size $n=2$ from a population of $N+8$ containing $N$ green balls, the expected value $E(X) = n \cdot \frac{N}{N+8}$.Given $E(X) = 1.2$, we have $2 \cdot \frac{N}{N+8} = 1.2$.$\frac{N}{N+8} = 0.6$.$N = 0.6(N+8) \implies N = 0.6N + 4.8$.$0.4N = 4.8 \implies N = \frac{4.8}{0.4} = 12$.

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$K + \frac{2}{10}$	$K + \frac{3}{10}$	$K + \frac{3}{10}$	$K + \frac{4}{10}$	$K + \frac{2}{10}$

$A$ box contains $8$ red and $N$ green balls. Two balls are drawn at random from it. If $X$ is the random variable representing the number of green balls drawn and $E(X) = 1.2$, then $N = ...$

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