CUET Preparation Today
A random variable X has the Probability Distribution :
The value of k is : |
$\frac{1}{3}$ $\frac{1}{5}$ $\frac{1}{2}$ $\frac{1}{6}$ |
$\frac{1}{3}$ |
The correct answer is Option (1) → $\frac{1}{3}$ The sum of all probabilities must be 1. $⇒2k^2+k+k+k^2=1$ $⇒3k^2+2k-1=0$ $⇒(k+1)(k-\frac{1}{3})-1=0$ $⇒k=\frac{1}{3}$ (Probability is always positive) |
