1 1 vote Let $X$ be a random variable with distribution function $F$, such that $$ \lim _{h \rightarrow 0^{-}} F(3+h)=\frac{1}{4} \text { and } F(3)=\frac{3}{4} . $$ Then $16 \operatorname{Pr}(X=3)$ equals $\qquad$ Probability goclasses_da_probability_wq7 numerical-answers goclasses one-mark probability + – GO Classes 285 views answer comment Share Follow Print See all 2 Comments 2 2 Comments reply Kd-here commented Jan 4 reply Follow flag Is this reasoning correct? F(3) = P(X<=3) P(X<=3) = P(X=3) + P(X<3) 0 0 replyShare Md._Wazir_Ali commented Aug 2 reply Follow flag Hi, How can the cdf at x = 3+h be lesser than CDF at x = 3? Can anyone explain this? We know that probability is positive. I think there is something wrong with the question. Thanks 0 0 replyShare Please log in or register to add a comment.
0 0 votes We are given:$\lim _{h \rightarrow 0^{-}} F(3+h)=\frac{1}{4}$$F(3)=\frac{3}{4}$Recall that for any distribution function $F(x)$, the probability that $X=x$ is given by the jump in the CDF at $x$ :$$\operatorname{Pr}(X=3)=F(3)-\lim _{h \rightarrow 0^{-}} F(3+h)$$Substitute the given values:$$\operatorname{Pr}(X=3)=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}$$So:$$16 \cdot \operatorname{Pr}(X=3)=16 \cdot \frac{1}{2}=8$$ GO Classes answered Jun 27, 2025 • edited Jul 1, 2025 by GO Classes GO Classes comment Share Follow 0 reply Please log in or register to add a comment.
