Probability of one discrete random variable having an higher value than another

Question

I am trying to understand a solved question about statistics.

I have two identical independent binomial random variables X and Y with identical pmfs

$P[X=x]={100 \choose x}0.05^x0.95^{100-x}$

and

$P[Y=y]={100 \choose y}0.05^y0.95^{100-y}$

Then, to calculate the probability of $A=\{X>5 , Y>3\}$

It says

$P[X >5, Y>3]= (1-P[X\leq 5])(1-P[Y\leq 3])$

I cannot visualize this. $(1-P[X\leq 5])$ is the probability of $X$ being greater than 5 and $(1-P[Y\leq 3])$ is the probability of $Y$ being greater than 3, thus, if I get it correctly, I will be calculating $P[X >5\cup Y>3]$ and not $P[X >5\cap Y>3]$, which is what it is asked. Can someone please tell me what am I understanding wrong?

I know that this is actually what has to be done, since it coincides with $F(\infty, \infty)-F(5, \infty)-F(\infty, 3)+F(5,3)$, but it seems to me that I am calculating the union, not the intersection.

Answer 1

Use independence first and then go to complements.

By independence $P[X>5,Y>3]=P[X>5] P[Y>3]$.

Also, $P[X>5]=1-P[X \leq 5]$ and $P[Y>3]=1-P[Y \leq 3]$.

So $P[X>5,Y>3]=(1-P[X \leq 5])(1-P[Y \leq 3])$.

Answer 2

Given any event $E$, let $p(E)$ denote the probability of event $E$ occurring.

Suppose that you have two independent events, $E_1$ and $E_2$.

Then, $p(E_1) \times p(E_2)$ represents the probability that events $E_1$ and $E_2$ both occur.

Further, the event that $E_1$ and $E_2$ both occur may be alternatively expressed as the event $(E_1 \cap E_2).$

As an example that is easy to visualize:

• Let $E_1$ be the event that a coin toss comes up Heads.
• Let $E_2$ be the event that the roll of a $6$ sided die comes up either $(1)$ or $(2)$.
• Assume that events $E_1$ and $E_2$ are independent events.

Then, the probability of events $E_1$ and $E_2$ both occurring, which may be expressed as $p(E_1 \cap E_2)$ is
$\frac{1}{2} \times \frac{1}{3}.$

Here, the probability that at least one of events $E_1$ and $E_2$ occurs, which may be expressed as $p(E_1 \cup E_2)$ is

$$ \left[\frac{1}{2} \times \frac{1}{3}\right] + \left[\frac{1}{2} \times \frac{2}{3}\right] + \left[\frac{1}{2} \times \frac{1}{3}\right].$$