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].

Hey Ðƴŋâmïč̣, good to see you again. Step 1: Find the value of K. The sum of all probabilities in a probability distribution must equal 1. P(Y) = 1 P(Y=0) + P(Y=1) + P(Y=2) + P(Y=3) + P(Y=4) = 1 (1)/(8) + 2K + (1)/(6) + 3K + (2)/(8) = 1 Combine the constant terms and the terms with K: ((1)/(8) + (2)/(8)) + (1)/(6) + (2K + 3K) = 1 (3)/(8) + (1)/(6) + 5K = 1 Find a common denominator for (3)/(8) and (1)/(6), which is 24: (3 × 3)/(8 × 3) + (1 × 4)/(6 × 4) + 5K = 1 (9)/(24) + (4)/(24) + 5K = 1 (13)/(24) + 5K = 1 Subtract (13)/(24) from both sides: 5K = 1 - (13)/(24) 5K = (24)/(24) - (13)/(24) 5K = (11)/(24) Divide by 5: K = (11)/(24 × 5) K = (11)/(120) The value of K is (11)/(120). Step 2: Find E[2Y + 4]. First, calculate the expected value of Y, E[Y]. E[Y] = Y · P(Y) E[Y] = (0 · (1)/(8)) + (1 · 2K) + (2 · (1)/(6)) + (3 · 3K) + (4 · (2)/(8)) E[Y] = 0 + 2K + (2)/(6) + 9K + (8)/(8) E[Y] = 11K + (1)/(3) + 1 E[Y] = 11K + (4)/(3) Substitute the value of K = (11)/(120): E[Y] = 11 (11