question_answer 1)

A random variable X has the following probability distribution : Determine : (i) K (ii) P (X < 3) (iii) P (X > 6) (iv) P (0 < X < 3)  

Answer:

Since       Therefore, 0 + K + 2K + 2K + 3K + K2 + 2K2 + K =  1        9 K + 10K2 =1                      10 K2 + 9K ?1 = 0        10 K2 + 10K ? K ?1 = 0        (K + 1) (10K 2 1)  = 0        10 K (K + 1) ? 1 (K + 1) = 0                         K is non-negative)       (i)         (ii)   P(X < 3)       = P (X = 0) + (X = 1) + P (X = 2)       (iii) P (X > 6) = P(X = 7) = 7K2 + K (i)     P(0 < X < 3) = (X = 1) + P(X = 2)