Probability of Y=0

By the principle of Inclusion and Exclusion,

$P(Y=0) = P(max(X,Y)=0) + P(min(X,Y)=0) - P(X=0)$ , hence this gives $p = 0.23 + 0.19 - 0.15 = 0.27$ , so the answer is $270$ .

Solution 1: The principle of Inclusion and Exclusion states that for any 2 events, we have

$P(A) + P (B) = P ( A \cap B ) + P ( A \cup B ).$

Let $A = \max(X, Y) = 0$ and $B = \min (X, Y) = 0$ . Then, we have

$P(\max(X,Y)=0) + P(\min(X,Y)=0) \\ = P ( \max(Y,Y) = 0 \cap \min (X, Y) = 0 ) + P ( \max(Y,Y) = 0 \cup \min (X, Y) = 0 ) \\ = P ( X = 0 \cap Y = 0 ) + P ( X = 0 \cup Y = 0).$

Now, let $A = ( X= 0 )$ and $B = ( Y = 0 )$ , and we get that

$P ( X = 0 ) + P ( Y = 0 ) = P ( X = 0 \cap Y = 0 ) + P ( X = 0 \cup Y = 0 ).$

Thus, we get that

$P(\max(X,Y)=0) + P(\min(X,Y)=0) \\ = P ( X = 0 \cap Y = 0 ) + P ( X = 0 \cup Y = 0) \\ = P ( X = 0 ) + P ( Y = 0 )$

Hence, $P(Y=0) = P(\max(X,Y)=0) + P(\min(X,Y)=0) - P(X=0)$ , hence this gives $p = 0.23 + 0.19 - 0.15 = 0.270$ , so the answer is $270$ .

Solution 2: To understand what $\max(X,Y)=0$ means, let's consider how this is possible. We clearly need $X \leq 0$ and $Y\leq 0$ . Furthermore, if we have $X < 0$ (respectively $Y<0$ ), then we must have $Y=0$ (resp. $X=0$ ). Adding on the case that $X=0, Y=0$ , we get that $P( \max (X, Y) = 0 ) = P( X =0, Y < 0) + P(X < 0, Y=0) + P(X=0, Y=0)$ . Applying the independence of the random variables, we get $P(\max (X, Y) = 0) = P(X=0) \cdot P(Y<0) + P(X<0) \cdot P(Y=0) + P(X=0) \cdot P(Y=0)$ .

Similarly, $\min(X,Y)=0$ breaks up into the cases $X>0, Y=0$ and $X=0, Y>0$ and $X=0, Y=0$ . Hence, we get that $P (\min(X,Y)=0) = P(X=0) \cdot P(Y>0) + P(X>0) \cdot P(Y=0) + P(X=0) \cdot P(Y=0)$ .

Adding up these two equations, we get that

$\begin{aligned} && P(\max(X,Y)=0) & = P(X=0) \cdot [ P(Y<0) + P(Y>0) ] + P(Y=0)[ P(X<0)+P(X>0)] \\ &+ & P(\min (X,Y)=0)) & \quad + 2 P(X=0)\cdot P(Y=0)\\ && & = P(X=0) \cdot [1-P(Y=0)] + P(Y=0) \cdot [1-P(X=0)] + 2 P(X=0) \cdot P(Y=0) \\ && & = P(X=0) + P(Y=0) \\ \end{aligned}$ where the identities $P(X<0)+P(X=0) + P(X>0)=1$ and $P(Y<0)+P(Y=0)+P(Y>0)=1$ are used in the second line. Thus, $0.23 + 0.19 = 0.15 + P(Y=0)$ , so $\lfloor 1000 P(Y=0) \rfloor= 270$ .

Work on each case of X,Y=0, negative or positive number.

Case 1: P(X=0, Y=0) = 0.15 \times p

Case 2: P(X=pos., Y=0) = \­(P(X=pos.) \times p\­)

Case 3: P(X=neg., Y=0) = \­(P(X=neg.) \times p\­)

Case 4: P(X=0, Y=pos.) = \­(0.15 \times P(Y=pos.)\­)

Case 5: P(X=0, Y=neg.) = \­(0.15 \times P(Y=neg.)\­)

Note that Cases 1, 3, 5 satisfy P(max(X,Y)=0) and Cases 1, 2, 4 satisfy P(min(X,Y)=0).

Solving linear equation:

\­(0.15p + P(X=neg.)p + 0.15P(Y=neg.)\­) = 0.23

\­(0.15p + P(X=pos.)p + 0.15P(Y=pos.)\­) = 0.19

Simplifying,

\­(0.15(P(Y=neg.) + P(Y=pos.) + 2p) + p(P(X=neg.) + P(X=pos.))\­) = 0.42

Further, \­(P(N=neg.) + P(N=pos.) = P(1-N), so:

\­(0.15(P(1-p) + 2p) + p(1-P(X))\­) = 0.42

\­(0.15 + 0.15p + p - 0.15p\­) = 0.42

\­(0.15 + p\­) = 0.42

p = 0.27

1000p = 270

p= P(Y=0)

p= [P(max(X,Y)=0) + P(min(X,Y)=0)] - P(X=0)

p = (0.23 + 0.19) - 0.15

p = 0.27

1000p = 270

p=0.19+0.23-0.15=0.27 1000p=0.27*1000+=270

The idea behind this solution is to think of each of the events $(X=min(X,Y))$ and $(Y=max(X,Y))$ as contrary to the event $(X=max(X,Y))$ . If events $A$ and $B$ are contrary, then $P(A)=1-P(B)$ . It basically means that if one of them happens, the other will not. And if one doesn't happen, the other will.

For easier redaction, I will state the following: $(X=max(X,Y))=(Xmax)$ , $(X=min(X,Y))=(Xmin)$ , $(Y=max(X,Y))=(Ymax)$ , $(Y=min(X,Y))=(Ymin)$ .

Firstly, $(max(X,Y)=0)$ is equivalent to ( $X$ being at the same time $max(X,Y)$ and equal to zero, or $Y$ being at the same time $max(X,Y)$ and equal to zero).

$P(max(X,Y)=0)=0.23 \Rightarrow$ $P(Xmax) \times P(X=0)+P(Ymax) \times P(Y=0)=0.23 \Rightarrow$ $P(Xmax) \times 0.15 + (1-P(Xmax)) \times P(Y=0)=0.23$ (1).

Secondly, $(min(X,Y)=0)$ is equivalent to ( $X$ being at the same time $min(X,Y)$ and equal to zero, or $Y$ being at the same time $min(X,Y)$ and equal to zero).

$P(min(X,Y)=0)=0.19 \Rightarrow$ $P(Xmin) \times P(X=0) + P(Ymin) \times P(Y=0)=0.19 \Rightarrow$ $(1-P(Xmax)) \times 0.15 + P(Xmax) \times P(Y=0)=0.19 \Rightarrow$ $0.15 - 0.15 \times P(Xmax) + P(Xmax) \times P(Y=0)=0.19$ (2).

$(1)+(2) \Rightarrow$ $P(Xmax) \times 0.15 + (1-P(Xmax)) \times P(Y=0) + 0.15 -$ $-0.15 \times P(Xmax) + P(Xmax) \times P(Y=0)=0.23+0.19 \Rightarrow$ $P(Xmax) \times 0.15 - 0.15 \times P(Xmax) + P(Y=0) \times (1-P(Xmax) +$ $+P(Xmax)) + 0.15 = 0.42 \Rightarrow$ $0 + P(Y=0) \times 1 = 0.42 - 0.15 \Rightarrow$ $P(Y=0) = 0.27 \Rightarrow$ $p=0.27 \Rightarrow$ $1000 \times p = 270$ .