A random variable X has a given distribution with P(X=k) involving a constant K.

Question

A random variable $X$ has a given distribution with $P(X=k)$ involving a constant $K$. Find $P(X>2)$.

Accepted Answer

Solving $6K^2+5K-1=0 \Rightarrow (6K-1)(K+1)=0 \Rightarrow K=\dfrac16$.

$$P(X>2) = K+2K+5K^2 = \frac16+\frac26+\frac{5}{36} = \frac{6+12+5}{36}=\frac{23}{36}$$

Solutions · 1