A random variable X has following probability function.X = xi01234567P

1.	A random variable X has following probability function.X = xi01234567P(xi)0k2k2k3kk22k27k2 + kThen find P(x < 2)
Answer» Since, sum of all probability of a probability distribution is 1 So, P(x = 0) + P(x = 1) + P(x = 2) + ... + P(x = 7) = 1 = 0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k =1 = 10k² + 9k = 0 = (10k - 1)(k + 1) = 0 ∴ k = 1/10, -1 but, k > 0 ∴ k = 1/10 Hence, P(x < 2) = P(x = 0) + P(x = 1) = 0 + k k = 1/10

A random variable X has following probability function.X = xi01234567P(xi)0k2k2k3kk22k27k2 + kThen find P(x < 2)

Answer»

Since, sum of all probability of a probability distribution is 1

So, P(x = 0) + P(x = 1) + P(x = 2) + ... + P(x = 7) = 1

= 0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k =1

= 10k² + 9k = 0

= (10k - 1)(k + 1) = 0

∴ k = 1/10, -1 but, k > 0

∴ k = 1/10

Hence, P(x < 2) = P(x = 0) + P(x = 1) = 0 + k

k = 1/10