1 1 vote Consider a discrete random variable $X$ with the following probability mass function (pmf):$$P(X=x)= \begin{cases}\frac{1}{4} & \text { if } x=2 \\\\ \frac{1}{2} & \text { if } x=4 \\\\ \frac{1}{4} & \text { if } x=6\end{cases}$$What is the value of $P\left(X^{2}-12 X+35>0\right)$ ?$\frac{1}{4}$ $\frac{3}{8}$ $\frac{1}{2}$ $\frac{3}{4}$ Probability goclasses goclasses-da-dpp goclasses-da-dpp-day-150 probability goclasses-probability-practice-questions random-variable + – GO Classes 148 views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
1 1 vote $X^2 - 12X + 35 = (X-5)(X-7)$.So we need to find when $(X-5)(X-7) > 0$.This is positive when $X < 5$ or $X > 7$.Now the possible values of $X$ are $2, 4, 6$.For $X=2$, $(2-5)(2-7) > 0$, so it works. For $X=4$, $(4-5)(4-7) > 0$, so it works. For $X=6$, $(6-5)(6-7) < 0$, so it does not work. Therefore,$P(X^2 - 12X + 35 > 0) = P(X=2) + P(X=4) = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$.So the correct answer is $\boxed{\frac{3}{4}}$. GO Classes answered Apr 17 GO Classes comment Share Follow 0 reply Please log in or register to add a comment.
