# Probability Function

• Oct 23rd 2008, 03:16 AM
da
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.
• Oct 23rd 2008, 04:02 AM
mr fantastic
Quote:

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) ....
• Oct 23rd 2008, 01:16 PM
Plato
Here is a bit more.
Quote:

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.

Quote:

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.
• Oct 23rd 2008, 02:53 PM
da
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?
• Oct 23rd 2008, 03:18 PM
Plato
I sorry that mislead you on the replacement case.
X=2 happens if (0,2),(2,0),(1,2),(2,1),(2,2) so five ways.
• Oct 23rd 2008, 04:05 PM
da
Ok cool. Thanks.

Is it true then with replacement Y has values 1,2,...,18. as {9,9} is a possible pick too?
• Oct 23rd 2008, 04:07 PM
da`
sorry 0,1,2,...,18 as {0,0} is also a possible pick?