Probability using the distributive law of a random variable

[filip_go]

Dear all,

After hours of hard work, I didn't manage to find the solution of this problem so I would really appreciate if you could help me solve this probability problem which is very useful in order to provide the sufficient points to pass an exam.

I would be very thankful if any of you could spare some of your precious time and post the solution until the 28th November.


There’s a standard deck of 52 cards. The ranks of the cards are 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K and А. Three cards are pulled from the pack without being returned.
Determine the distributive law of the random variable X that presents the absolute value of the subtraction between the ranks of the first and the second chosen cards.
Using the random variable X, express and calculate the probability that the rank of the third card which will be pulled, will be between the ranks of the two previously pulled cards. For example, if the ranks of the first two cards are 4 and 9, determine the probability that the rank of the third card will be 5, 6, 7 or 8.
(The ranks and the sequence are already given. The rank of the Jack is 11, and the rank of the Ace is 14. )

[Guest]

[tex]X=0,1,2,3,4,5,6,7,8,9,10,11,12[/tex]
[tex]P(X=0)=\frac{13 {4 \choose 2}}{{52 \choose 2}}[/tex]

[tex]P(X=k)=\frac{(13-k).{4 \choose 1}.{4 \choose 1}}{{52 \choose 2}}, \ \ k=1,2,3,4,5,6,7,8,10,11,12[/tex]

[kmitov]

Let us denote by A the event that the third number is between the first and the second one.
Then we have
[tex]P(A|X=0)=0, P(A|X=1)=0[/tex]

[tex]P(A|X=2)=\frac{4}{50}[/tex]
[tex]P(A|X=3)=\frac{8}{50}[/tex]
[tex]P(A|X=4)=\frac{12}{50}[/tex]
[tex]P(A|X=5)=\frac{16}{50}[/tex]
[tex]P(A|X=6)=\frac{20}{50}[/tex]
[tex]P(A|X=7)=\frac{24}{50}[/tex]
[tex]P(A|X=8)=\frac{28}{50}[/tex]
[tex]P(A|X=9)=\frac{32}{50}[/tex]
[tex]P(A|X=10)=\frac{36}{50}[/tex]
[tex]P(A|X=11)=\frac{40}{50}[/tex]
[tex]P(A|X=12)=\frac{44}{50}[/tex]
Now using the distribution of X, we have to apply the total probability formula.