3.39 From a sack of fruit containing 3 oranges, 2 apples, and 3 bananas, a random sample of 4 pieces of fruit is selected. If X is the number of oranges and Y is the number of apples in the sample, find (a) the joint probability distribution of X and Y ; (b) P[(X, Y ) ∈ A], where A is the region that is given by {(x, y) | x + y ≤ 2}.

Step-by-step explanation:Let's start defining the random variables.X : ''The number of oranges in the sample of 4 fruits''Y : ''The number of apples in the sample of 4 fruits''X can assume values 0,1,2,3 because they are 3 oranges in the sack of fruit.Y can assume values 0,1,2 because they are 2 apples in the sack of fruit.But for the domain of the joint probability distribution [tex](X=3,Y=2)[/tex] it is not a possible value because this will mean that the sample contains 5 fruits.Also [tex](X=0,Y=0)[/tex] it is an impossible value because given that we selected the 3 bananas we must complete the sample with an orange or either an apple.We conclude that [tex]P(X=0,Y=0)=P(X=3,Y=2)=0[/tex]There is not another restriction for the domain of [tex](X=x,Y=y)[/tex]Let's define the combinatorial number.[tex]nCr=\frac{n!}{r!(n-r)!}[/tex]The joint probability distribution of X and Y is :[tex]f(x,y)=P(X=x,Y=y)=\frac{(3Cx)(2Cy)[3C(4-x-y)]}{(8C4)}[/tex]From 3 oranges I selected ''x'' oranges.From 2 apples I selected ''y'' apples.From 3 bananas I selected ''4-x-y'' bananas that is the total amount of bananas minus the number of oranges and apples.We multiply this three combinatorial numbers and then we divide by the total forms to choose 4 fruits between a total of [tex]3+2+3=8[/tex] fruits.Now we listed all the values of [tex]P(X=x,Y=y)[/tex] :[tex]P(X=0,Y=0)=0[/tex][tex]P(X=1,Y=0)=\frac{3}{70}[/tex][tex]P(X=2,Y=0)=\frac{9}{70}[/tex][tex]P(X=3,Y=0)=\frac{3}{70}[/tex][tex]P(X=0,Y=1)=\frac{2}{70}[/tex][tex]P(X=1,Y=1)=\frac{18}{70}[/tex][tex]P(X=2,Y=1)=\frac{18}{70}[/tex][tex]P(X=3,Y=1)=\frac{2}{70}[/tex][tex]P(X=0,Y=2)=\frac{3}{70}[/tex][tex]P(X=1,Y=2)=\frac{9}{70}[/tex][tex]P(X=2,Y=2)=\frac{3}{70}[/tex][tex]P(X=3,Y=2)=0[/tex]b) P [(X,Y) ∈ A] where A is the region that is given by {(x, y) | x + y ≤ 2} is equal to the sum of the probability for the (X,Y) points that satisfy A.P [(X,Y) ∈ A] = [tex]P(X=0,Y=0)+P(X=1,Y=0)+P(X=2,Y=0)+P(X=1,Y=1)+P(X=0,Y=1)+P(X=0,Y=2)[/tex]P [(X,Y) ∈ A] = [tex]0+\frac{3}{70}+\frac{9}{70}+\frac{18}{70}+\frac{2}{70}+\frac{3}{70}=\frac{35}{70}=\frac{1}{2}=0.5[/tex]