0 votes 0 votes Let $X$ and $Y$ be independent discrete random variables with probability mass functions $P(X=$ $k)$ and $P(Y=k)$.What is $P(\min \{X, Y\} \leq x)$ ? $P(\min \{X, Y\} \leq x)=1-P(X>x) P(Y>x)$ $P(\min \{X, Y\} \leq x)=P(X \leq x) P(Y \leq x)$ $P(\min \{X, Y\} \leq x)=P(X \leq x)+P(Y \leq x)-P(X \leq x) P(Y \leq x)$ $P(\min \{X, Y\} \leq x)=P(X>x)+P(Y>x)-P(X>x) P(Y>x)$ Probability goclasses2025-da-probability-3 goclasses probability 2-marks + – GO Classes asked Jun 5 • edited Jun 12 by GO Classes GO Classes 92 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes $$ \begin{aligned} P(\min \{X, Y\} \leq x)= & 1-P(\min \{X, Y\}>x) \\ & =1-P(X>x \text { and } Y>x) \\ & =1-P(X>x) \cdot P(Y>x) \\ & =1-P(X>x) P(Y>x) \end{aligned} $$ GO Classes answered Jun 5 GO Classes comment Share Follow See all 3 Comments See all 3 3 Comments reply Psrc_2025 commented Jun 6 reply Follow Share The answer should be option C which is P(X<=x or Y<=y) the relation should be or not and as min (X,Y) <=x doesnt imply that max (X,Y) should also be <=x . So both X and Y need not be <=x . 1 votes 1 votes Psrc_2025 commented Jun 6 reply Follow Share And also option A and option C are equivalent ..one can be reduced to another 0 votes 0 votes ankhu1610 commented Jun 13 reply Follow Share option a and c both are true 0 votes 0 votes Please log in or register to add a comment.
