Question

A random variable ?X? has the following probability distribution: x 1 2 3 4 5 6 7 P(x) k-1 3k k 3k $3k^2$ $k^2$ $k^2 + k$ Then the value of k is

• A. $\frac{2}{7}$
• B. $\frac{1}{5}$
• C. $\frac{1}{10}$
• D. -2

Answer 1

Answer: (B)

$\frac{1}{5}$

Answer 2

$\sum P_{i}=1$
$k-1+3k+k+3k+3k^{2}+k^{2}+k^{2}+k=1$
$5k^{2}+9k-2=0$
$5k^{2}+10k-k-2=0$
$5k\left(k+2\right)-1\left(k+2\right)=0$
$\left(5k-1\right)\left(k+2\right)=0$
$k=\frac{1}{5},-2\left(k=-2\, is\, not\, possible\right).$