# Compute the probability of each of the following events

• December 15th 2009, 02:26 AM
zorro
Compute the probability of each of the following events
Question : Two discrete random variables X and Y have joint pmf given by the following table:
$
\begin{array}{c|c|c|c}.& Y=1 & Y=2 & Y=3 \\
\hline
X=1 & \frac{1}{12} & \frac{1}{6} & C \\
X=2 & \frac{1}{6} & \frac{1}{4} & \frac{1}{12} \\
X=3 & \frac{1}{12} & \frac{1}{12} & 0 \\
\end{array}$

i) Find C
ii) Compute the probability of each of the following events.
- X $\le$ 1 1/2
- X is odd
- XY is even
- Y is odd given that X is odd
• December 15th 2009, 08:00 AM
ANDS!
What are you having problems with specifically with this problem? What have you tried so far? For the first part its a simple application of the fact that given the sample space S, P(S)=1.

As for the second part, I would first compute the marginal probabilities of X and Y, and see if that helps you knock out some of those problems.
• December 17th 2009, 04:28 PM
zorro
the question i)
Quote:

Originally Posted by ANDS!
What are you having problems with specifically with this problem? What have you tried so far? For the first part its a simple application of the fact that given the sample space S, P(S)=1.

As for the second part, I would first compute the marginal probabilities of X and Y, and see if that helps you knock out some of those problems.

I am having problem with the first question i) Find C
• December 18th 2009, 10:36 AM
ANDS!
Remember that the probability that an outcome occurs in a sample space is 1 (or 100% chance that some event is going to happen). So if you know that the Probability of X and Y occurring MUST add up to 1, and you have all probabilities except for C, how do you go about figuring out what C is?
• December 18th 2009, 12:54 PM
zorro
I am still having problems
Quote:

Originally Posted by ANDS!
Remember that the probability that an outcome occurs in a sample space is 1 (or 100% chance that some event is going to happen). So if you know that the Probability of X and Y occurring MUST add up to 1, and you have all probabilities except for C, how do you go about figuring out what C is?

Here is what u have done till now....

C = $\frac{1}{12}$ ..........Is that correct?

Could u please explain me what does 'X is odd' mean, i tried searching on the net for the term odd random variable but couldnt find any meaning ...........please explain me when is a random variable termed odd or even
• December 18th 2009, 01:15 PM
ANDS!
Quote:

Originally Posted by zorro
Here is what u have done till now....

C = $\frac{1}{12}$ ..........Is that correct?

Could u please explain me what does 'X is odd' mean, i tried searching on the net for the term odd random variable but couldnt find any meaning ...........please explain me when is a random variable termed odd or even

The first part is correct. For the second part, it is literally asking you for the probability of X being 1 and 3 - odd numbers. No trick.
• December 18th 2009, 01:36 PM
zorro
So here what i have ?
Quote:

Originally Posted by ANDS!
The first part is correct. For the second part, it is literally asking you for the probability of X being 1 and 3 - odd numbers. No trick.

So its just asking me if the F(X) = odd or even

So since $F(x) = Pr(X \le x ) = 1$ .........therefore i guess it is odd.........right!!

And XY would be odd ................am i correct
How would i do the last one which says
Y is odd given that Xis odd
• December 18th 2009, 07:22 PM
ANDS!
I think you are confusing yourself with what the question is asking. Not to be blunt, but literally, you are being asked for the random variables whose values are odd (i.e. not even, not divisible by 2). If I asked you to toss a dice, and give me the probability of getting an ODD NUMBER, how would you go about that? This question is no different:

The probability of X being odd is P(X=1 or X=3).

The probability of the random variable XY being odd is P(XY=1, 3, 9).

The last one is a conditional. Do you know how to compute marginal probabilities? What is the probability of X being equal to 1 over ALL values of Y (since this is a joint pdf)? That is the probability of X being equal to 1. We call them "marginals", because you would (to make it less messy) scribble the total probabilities in the "margins" of your chart there. Once you compute the marginal, it is simple to compute P(Y|X is odd).
• December 18th 2009, 09:21 PM
zorro
Is this correct?
Quote:

Originally Posted by ANDS!
I think you are confusing yourself with what the question is asking. Not to be blunt, but literally, you are being asked for the random variables whose values are odd (i.e. not even, not divisible by 2). If I asked you to toss a dice, and give me the probability of getting an ODD NUMBER, how would you go about that? This question is no different:

The probability of X being odd is P(X=1 or X=3).

The probability of the random variable XY being odd is P(XY=1, 3, 9).

The last one is a conditional. Do you know how to compute marginal probabilities? What is the probability of X being equal to 1 over ALL values of Y (since this is a joint pdf)? That is the probability of X being equal to 1. We call them "marginals", because you would (to make it less messy) scribble the total probabilities in the "margins" of your chart there. Once you compute the marginal, it is simple to compute P(Y|X is odd).

$\therefore$ Pr(X is odd) = Pr(X=1)+Pr(X=3) = $\frac{1}{3} + \frac{1}{6}$ = $\frac{3}{6}$ = $\frac{1}{2}$.......Is this correct?

And
Pr(XY is even ) = Pr(XY=2) = $\frac{1}{2} + \frac{1}{2}$ = 1.........Is this correct?

Pr(X is odd | Y is odd) = $\frac{Pr(X \cup Y) }{Pr(Y)}$ = $\frac{\frac{1}{3} . \frac{1}{6}}{\frac{1}{2}}$.......Is this correct ?
• December 20th 2009, 07:18 PM
ANDS!
Quote:

Originally Posted by zorro
$\therefore$ Pr(X is odd) = Pr(X=1)+Pr(X=3) = $\frac{1}{3} + \frac{1}{6}$ = $\frac{3}{6}$ = $\frac{1}{2}$.......Is this correct?Yes.

And
Pr(XY is even ) = Pr(XY=2) = $\frac{1}{2} + \frac{1}{2}$ = 1.........Is this correct? No. You should be able to check why this is not correct. The probability that XY is equal to is the P(1,2)+P(2,1). That is equal to 1/6 (from your table above). You should also know, from the fundamentals of probability, that those two answers can NOT add up to 1, since that would mean that the probability of any other XY is 0, which is not true from your table. Go slow, and figure out all the possible combinations of X multiplied by Y that will give you an even number. Your possible values of X are 1, 2 and 3 and your possilbe values of Y are 1, 2 and 3.

Pr(X is odd | Y is odd) = $\frac{Pr(X \cup Y) }{Pr(Y)}$ = $\frac{\frac{1}{3} . \frac{1}{6}}{\frac{1}{2}}$.......Is this correct ?No. The probability that Y is odd is not 1/2. It is P(Y=1,3). That probability is 2/3 - calculate that using the marginal probabilities. You never responded if you knew how to calculate marginals. The top of your quotient isn't correct either. It is the P(X AND Y are odd), not X or Y. If it were X or Y, the probability would be 1 on the top. Which is impossible.