CUET Preparation Today
Probability distribution of random variable X is
The value of k is: |
$k=-1, \frac{1}{9}$ $k=-1$ $k=\frac{1}{9}$ $k=0,-9$ |
$k=\frac{1}{9}$ |
The correct answer is Option (3) - $k=\frac{1}{9}$ $∑P(x_i)=1$ $3k+3k+2k+2k^2+3k^2+4k^2=1$ $8k+9k^2-1=0$ $9k^2+9k-k-1=0$ $9k(k+1)-1(k+1)=0$ $k=\frac{1}{9}$ |
