Question

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)

X	0	1	2	3	4	5	6	7
P(X)	0	K	2K	2K	3K	K2	2K2	7K2+K

Answer 1

(i) It is known that the sum of probabilities of a probability distribution of random variables is one.
∴P(X=0)+P(X=1)+....+P(X=7)=1
⇒0+k+2k+2k+3k+k²+2k²+(7k²+k)=1
⇒10k²+9k-1=0
⇒(10k-1)(k+1)=0
⇒k=-\(\frac{1}{10}\) or k=-1
Since,k≥0,therefore k=-1 is not possible.
\(∴k=\frac{1}{10}\)
(ii) P (X < 3) = P (X = 0) + P (X = 1) + P (X = 2)
=0+k+2k
\(=3k=3×\frac{1}{10}=\frac{3}{10}\)
(iii) P (X > 6) = P (X = 7)
=7k²+k=\(7(\frac{1}{10})^2\)+\(\frac{1}{10}\)=\(\frac{17}{100}\)
(iv) P (0 < X < 3) = P (X = 1) + P (X = 2)
=k+2k=3k

\(=3X\frac{3}{10}\)

= \(\frac{3}{10}\)