Discrete Random Variables

[Mo]

Hello (first time poster),
i am having quite a bit of trouble with a particular problem on stats (which i despise of!) - in particular, discrete random variables.Ok here is the question:


"Find the probability distribution of X in each of the following questions ...

Two fair dice are thrown. X is the smaller of the two scores on the dice"


I have managed to do the first few, but this seems to have thrown me a little bit.I would be grateful for any help indeed.If anyone knows of any good resources online for AS-level Stats, i would also be grateful for the link.

Regards,
Mo

[nnnnnnnn]

Is that a question or part of what is given?

If you want the probability that a particular die is the smaller of two, you would just have to count all the possibilities for the dice not being equal and divide by 36. There are 6 cases that the dice are equal, this leaves 30. You only want half of these because in 15 cases X is smaller, in the other 15, the other die is smaller. So it is 15/36.

[Galileo]

"Find the probability distribution of X in each of the following questions ...

Two fair dice are thrown. X is the smaller of the two scores on the dice"

Hi.

It's not too hard to find P(X=a) for a certain a. You just count all the outcomes which have a as a lowest score and divide by 36.

For example: P(X=3)=7/36.
(7 outcomes: (3,3),(3,4),(3,5),(3,6),(4,3),(5,3),(6,3))

[Mo]

That is a whole question (well, sub-question) Believe it or not, i think i almost had it. here was my working out for the question:

X = Smaller of two scores

SET =
1,1 1,2 1,3 1,4 1,5 1,6
2,1 2,2 2,3 2,4 2,5 2,6
3,1 3,2 3,3 3,4 3,5 3,6
4,1 4,2 4,3 4,4 4,5 4,6
5,1 5,2 5,3 5,4 5,5 5,6
6,1 6,2 6,3 6,4 6,5 6,6

i then carried on with ..

P (X=1) P(1, any other except for 1) =5/36
P (X=2) P(2, any other except for 1 or 2) =4/36
P (X=3) P(3, any other except for 1,2 or 3) =3/36
P (X=4) P(4, 5 or 6) =2/36
P (X=1) P(5, 6) =1/36

Where i was getting confused was because i had been taught that the probabilities should add always add upto 1?

Thanks for your help so far :)

[nnnnnnnn]

You did have it... you just need to add up your P(x=1)+...+P(x=5) to get 15/36.

P(X is smaller) + P(X isnt smaller) = 1 = 100%, thats when it adds to 1...

[Mo]

Ahh, yes, as in the total probabaility will be equal to 1.Thanks for your help, it is much appreciated.

[nnnnnnnn]

Hi.

It's not too hard to find P(X=a) for a certain a. You just count all the outcomes which have a as a lowest score and divide by 36.

For example: P(X=3)=7/36.
(7 outcomes: (3,3),(3,4),(3,5),(3,6),(4,3),(5,3),(6,3))

He needs it for a particular die and equal shouldn't be counted, so P(X=3) = 3/36

[Galileo]

He needs it for a particular die and equal shouldn't be counted, so P(X=3) = 3/36
That's not what the question asks.

A random variable is a function from the sample space to [0,1].
If the event is throwing a 3 and a 6 - (3,6), then X=3. If the event is (2,4), then X=2.
And ofcourse (3,3) gives X=3.
The probability of getting X=3 is 7/36 because of the reason I previously gave.

He is asked for the distrubution function, which according to my definition is:
F(a)=P(X\leq a)
although I think the term is easily confused with the probability mass function:
p(a)=P(X=a)

[nnnnnnnn]

That's not what the question asks.

A random variable is a function from the sample space to [0,1].
If the event is throwing a 3 and a 6 - (3,6), then X=3. If the event is (2,4), then X=2.
And ofcourse (3,3) gives X=3.
The probability of getting X=3 is 7/36 because of the reason I previously gave.

He is asked for the distrubution function, which according to my definition is:
F(a)=P(X\leq a)
although I think the term is easily confused with the probability mass function:
p(a)=P(X=a)
I didn't realize that probability distribution was a function, guess I shouldnt have answered...

He thought it should have added to 1, your way does so thats probably why he thought that. Sorry Mo...

[Mo]

Thanks for your help both.I completed the answer below just for anyone's future reference:


X = Smaller of two scores

SET =
1,1 1,2 1,3 1,4 1,5 1,6
2,1 2,2 2,3 2,4 2,5 2,6
3,1 3,2 3,3 3,4 3,5 3,6
4,1 4,2 4,3 4,4 4,5 4,6
5,1 5,2 5,3 5,4 5,5 5,6
6,1 6,2 6,3 6,4 6,5 6,6

(another, most probabaly easier method of viewing would be ...)

1 2 3 4 5 6
1 1 1 1 1 1 1
2 1 2 2 2 2 2
3 1 2 3 3 3 3
4 1 2 3 4 4 4
5 1 2 3 4 5 5
6 1 2 3 4 5 6

(basically this shows which number is the smallest out of the pair (from both die))

i then carried on with ..

P (X=1) P(where 1 is smallest number) =11/36
P (X=2) P(where 2 is smallest number) =9/36
P (X=3) P(where 3 is smallest number) =7/36
P (X=4) P(where 4 is smallest number =5/36
P (X=5) P(where 5 is smallest number) =3/36
P (X=6) P(where 6 is smallest number) =1/36

Notice how they do all add up to 1 (36/36)

Regards,
Mo