1. ## Probability Function

Struggling here with this question would really appreciate and be very thankful for any help. It seems really general to me and clearly don't have a knack for this, and just need to understand how create a probability function.

Quote:
Two cards are picked from a bag containing 10 cards numbered 0,1,2,.....,9. Let X be the larger number of the two cards and let Y be their total.

Find the probability function of X and the probability function of Y, if sampling is without replacement.

Find the probability function of X and the probability function of Y, if sampling is with replacement.

2. Originally Posted by da
Struggling here with this question would really appreciate and be very thankful for any help. It seems really general to me and clearly don't have a knack for this, and just need to understand how create a probability function.

Quote:
Two cards are picked from a bag containing 10 cards numbered 0,1,2,.....,9. Let X be the larger number of the two cards and let Y be their total.

Find the probability function of X and the probability function of Y, if sampling is without replacement.

Find the probability function of X and the probability function of Y, if sampling is with replacement.
Some things to think about when finding the distribution for X:

What event gives X = 0? So what's Pr(X = 0) ....
What events give X = 1? So what's Pr(X = 1) ....
What events give X = 2? So what's Pr(X = 2) ....
.
.
.
What events give X = 9? So what's Pr(X = 9) ....

Things to think about when finding the distribution for Y:

What event gives Y = 0? So what's Pr(Y = 0) ....
What events give Y = 1? So what's Pr(Y = 1) ....
What events give Y = 2? So what's Pr(Y = 2) ....
.
.
.
What events give Y = 18? So what's Pr(Y = 18) ....

3. Here is a bit more.
Originally Posted by da
Two cards are picked from a bag containing 10 cards numbered 0,1,2,.....,9. Let X be the larger number of the two cards and let Y be their total.
Find the probability function of X and the probability function of Y, if sampling is without replacement.
Without replacement there are $\displaystyle {10 \choose 2}=45$ ways to choose two cards.
That is the denominator of the probabilities.
Now X can have 9 values: 1,2,…,9. $\displaystyle X=5$ in five ways: {0,5},{1,5},{2,5},{3,5},{4,5}.
Thus $\displaystyle P(X=5)=\frac{5}{45}$.
You do the others.

Now Y can have values 1,2,…,17. $\displaystyle Y=8$ in four ways: {0,8},{1,7},{2,6},{5,3}.
Thus $\displaystyle P(Y=8)=\frac{4}{45}$.
You do the others.

Originally Posted by da`
Two cards are picked from a bag containing 10 cards numbered 0,1,2,.....,9. Let X be the larger number of the two cards and let Y be their total.
Find the probability function of X and the probability function of Y, if sampling is with replacement.
With replacement there are $\displaystyle {10}^2 =100$ ways to choose two cards.
That is the denominator of the probabilities.
Now X can have 10 values: 0,1,2,…,9. $\displaystyle X=5$ in six ways: {0,5},{1,5},{2,5},{3,5},{4,5},{5,5}.
Thus $\displaystyle P(X=5)=\frac{6}{100}$.
You do the others.

Now Y can have values 1,2,…,17. $\displaystyle Y=8$ in five ways: {0,8},{1,7},{2,6},{5,3},{4,4}.
Thus $\displaystyle P(Y=8)=\frac{5}{100}$.
You do the others.

4. Thank you very much for your help. My apologies to the moderators for my re-posts.

Right Ive got the without replacement down fine but its just the with replacement one I am having trouble with.

Without

P(X=1) = 1/45
P(X=2) = 2/45
P(X=3) = 3/45
P(X=4) = 4/45
P(X=5) = 5/45
P(X=6) = 6/45
P(X=7) = 7/45
P(X=8) = 8/45
P(X=9) = 9/45

P(Y=1) = 1/45
P(Y=2) = 1/45
P(Y=3) = 2/45
P(Y=4) = 2/45
P(Y=5) = 3/45
P(Y=6) = 3/45
P(Y=7) = 4/45
P(Y=8) = 4/45
P(Y=9) = 5/45
P(Y=10) = 4/45
P(Y=11) = 4/45
P(Y=12) = 3/45
P(Y=13) = 3/45
P(Y=14) = 2/45
P(Y=15) = 2/45
P(Y=16) = 1/45
P(Y=17) = 1/45

With (ill explain my problem)

so 100 ways to choose two cards

P(X=0) = 1/100 eg {0,0}
P(X=1) = 2/100 eg {0,1}, {1,1}
P(X=2) = 3/100 eg {0,2}, {1,2}, {2,2}
P(X=3) = 4/100
P(X=4) = 5/100
P(X=5) = 6/100
P(X=6) = 7/100
P(X=7) = 8/100
P(X=8) = 9/100
P(X=9) = 10/100

(1+2+...+10)/100= 55/100

P(Y=1) = 1/100 eg {0,1}
P(Y=2) = 2/100 eg {0,2}, {1,1}
P(Y=3) = 2/100
P(Y=4) = 3/100
P(Y=5) = 3/100
P(Y=6) = 4/100 eg {1,5}, {0,6}, {2,4}, {3,3}
P(Y=7) = 4/100
P(Y=8) = 5/100
P(Y=9) = 5/100
P(Y=10) = 5/100
P(Y=11) = 4/100
P(Y=12) = 4/100
P(Y=13) = 3/100
P(Y=14) = 3/100
P(Y=15) = 2/100
P(Y=16) = 2/100
P(Y=17) = 1/100

(2x1) + (4x2) + (4x3) + (4x4) + (3x5) = 53/100

which is clearly wrong! what am i doing wrong?

5. I sorry that mislead you on the replacement case.