# Math Help - Finding the probability

1. ## Finding the probability

Let $P(X=i)=P(Y=i)=\frac{1}{2^i}$
Where $(i=1,2,...)$.
X and Y are independent.

Then how can I find $P(min(X,Y)\le i)$.

I know that when $P(max(X,Y)\le i)$ means that both X and Y are less than i so I can conclude that $P(X\le i,Y\le i)=P(X\le i)P(Y\le i)$. But how can I do the same for Min?

Meanwhile, since this is a discrete distribution, I am also facing the problem of finding its mass probability function. e.g. $P(X\le i)=\sum_{X_n to solve for some other parts. Any suggestion?

TIA

2. Originally Posted by noob mathematician
Let $P(X=i)=P(Y=i)=\frac{1}{2^i}$
Where $(i=1,2,...)$.
X and Y are independent.

Then how can I find $P(min(X,Y)\le i)$.

I know that when $P(max(X,Y)\le i)$ means that both X and Y are less than i so I can conclude that $P(X\le i,Y\le i)=P(X\le i)P(Y\le i)$. But how can I do the same for Min?

Meanwhile, since this is a discrete distribution, I am also facing the problem of finding its mass probability function. e.g. $P(X\le i)=\sum_{X_n to solve for some other parts. Any suggestion?

TIA

The equation I will use is: $P(X_{k}\le x)=\sum_{j=0}^{n-k}\dbinom{n}{j}\pi_3^j(1-\pi_3)^{n-j}$

Is it true that for this case, I'm finding:

$\sum_{j=0}^{2-1}\dbinom{2}{j}\pi_3^j(1-\pi_3)^{2-j}$

How do I find $\pi_3=P(X>x)$?

4. Originally Posted by noob mathematician

The equation I will use is: $P(X_{k}\le x)=\sum_{j=0}^{n-k}\dbinom{n}{j}\pi_3^j(1-\pi_3)^{n-j}$

Is it true that for this case, I'm finding:

$\sum_{j=0}^{2-1}\dbinom{2}{j}\pi_3^j(1-\pi_3)^{2-j}$

How do I find $\pi_3=P(X>x)$?
I would have thought that n = 2 and k = 2 ...

Also, $\displaystyle \Pr(X > i) = \frac{1}{2^{i+1}} + \frac{1}{2^{i+2}} + .... = \frac{1}{2^{i}} \left( \frac{1}{2} + \frac{1}{2^2} + .... \right)$ and you should know what the sum in the brackets is equal to.

5. Thanks.

Would like to extend my question.

What does it means by $P(\text{X divides Y})$ in this case?

Is it possible to generate its probability?