A random variable Y has the following probability distribution: P(Y=0)=1/8, P(Y=1)=2K, P(Y=2)=1/6, P(Y=3)=3K, P(Y=4)=2/8. Find the value of K and E[2Y+4].

Step 1: Find the value of $K$.
The sum of all probabilities in a probability distribution must equal 1.

$$\sum P(Y) = 1$$ $$P(Y=0) + P(Y=1) + P(Y=2) + P(Y=3) + P(Y=4) = 1$$ $$\frac{1}{8} + 2K + \frac{1}{6} + 3K + \frac{2}{8} = 1$$

Combine the constant terms and the terms with $K$:

$$\left(\frac{1}{8} + \frac{2}{8}\right) + \frac{1}{6} + (2K + 3K) = 1$$ $$\frac{3}{8} + \frac{1}{6} + 5K = 1$$

Find a common denominator for $\frac{3}{8}$ and $\frac{1}{6}$, which is 24:

$$\frac{3 \times 3}{8 \times 3} + \frac{1 \times 4}{6 \times 4} + 5K = 1$$ $$\frac{9}{24} + \frac{4}{24} + 5K = 1$$ $$\frac{13}{24} + 5K = 1$$

Subtract $\frac{13}{24}$ from both sides:

$$5K = 1 - \frac{13}{24}$$ $$5K = \frac{24}{24} - \frac{13}{24}$$ $$5K = \frac{11}{24}$$

Divide by 5:

$$K = \frac{11}{24 \times 5}$$ $$K = \frac{11}{120}$$

The value of $K$ is $\boxed{\frac{11}{120}}$.

Step 2: Find $E[2Y + 4]$.
First, calculate the expected value of $Y$, $E[Y]$.

$$E[Y] = \sum Y \cdot P(Y)$$ $$E[Y] = \left(0 \cdot \frac{1}{8}\right) + (1 \cdot 2K) + \left(2 \cdot \frac{1}{6}\right) + (3 \cdot 3K) + \left(4 \cdot \frac{2}{8}\right)$$ $$E[Y] = 0 + 2K + \frac{2}{6} + 9K + \frac{8}{8}$$ $$E[Y] = 11K + \frac{1}{3} + 1$$ $$E[Y] = 11K + \frac{4}{3}$$

Substitute the value of $K = \frac{11}{120}$:

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