Three cards are drawn without replacement from the 12 face cards (jack

ni3s4otqc

ni3s4otqc

Answered question

2021-11-28

Three cards are drawn without replacement from the 12 face cards (jacks, queens, and kings) of an ordinary deck of 52 playing cards. Let X be the number of kings selected and Y the number of jacks.
a) Find the joint probability distribution of X and Y;
b) Find the marginal probability distributions of X and Y.
c) P[(X,Y)A], where A is the region given by {(x,y)x+y2}.

Answer & Explanation

Novembruuo

Novembruuo

Beginner2021-11-29Added 26 answers

Step 1
Total number of cards =52
Total number of face cards =12
In a pack of 52 cards, there are total 4 Kings and 4 Jacks. It is given that 3 cards are drawn. So, the possible values of X and Y are: 0, 1, 2 and 3.
Now, the probability of drawing x kings and y Jacks can be calculated using the formula.
P(X=x,Y=y)=(4x)×(4y)×(43xy)(123)
After drawing x kings, y jacks, (3xy) would be left. Here, (x+y) should be always less than equal to 3.
Step 2
a) The joint probability distribution of X and Y:
Y/x0123Total Y01/556/556/551/5514/5516/5516/556/55028/5526/556/550012/5531/550001/55Total X14/5528/5512/551/551
Step 3
b) The marginal Probability distribution of X and y are
Marginal Distribution of X
XMarginal Distribution014/55128/55212/5531/55
Marginal distribution of Y
YMarginal Distribution014/55128/55212/5531/55
Step 4
c) To find: P[X,Y)A] where A is the region given by {(x,y)x+y2}
That is P(X,YA)=P(x+y2)
Now, P=1P(x+y1)
=1[[P(X=0,Y=0]+[P(X=0,Y=1]+[P(X=1,Y=0]]
=1[155+655+655]
=11355
=4255

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