Question

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}.

Answer 1

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 $(X=3,Y=2)$ it is not a possible value because this will mean that the sample contains 5 fruits.

Also $(X=0,Y=0)$ 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 $P(X=0,Y=0)=P(X=3,Y=2)=0$

There is not another restriction for the domain of $(X=x,Y=y)$

Let's define the combinatorial number.

$nCr=\frac{n!}{r!(n-r)!}$

The joint probability distribution of X and Y is :

$f(x,y)=P(X=x,Y=y)=\frac{(3Cx)(2Cy)[3C(4-x-y)]}{(8C4)}$

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 $3+2+3=8$ fruits.

Now we listed all the values of $P(X=x,Y=y)$ :

$P(X=0,Y=0)=0$

$P(X=1,Y=0)=\frac{3}{70}$

$P(X=2,Y=0)=\frac{9}{70}$

$P(X=3,Y=0)=\frac{3}{70}$

$P(X=0,Y=1)=\frac{2}{70}$

$P(X=1,Y=1)=\frac{18}{70}$

$P(X=2,Y=1)=\frac{18}{70}$

$P(X=3,Y=1)=\frac{2}{70}$

$P(X=0,Y=2)=\frac{3}{70}$

$P(X=1,Y=2)=\frac{9}{70}$

$P(X=2,Y=2)=\frac{3}{70}$

$P(X=3,Y=2)=0$

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] = $P(X=0,Y=0)+P(X=1,Y=0)+P(X=2,Y=0)+P(X=1,Y=1)+P(X=0,Y=1)+P(X=0,Y=2)$

P [(X,Y) ∈ A] = $0+\frac{3}{70}+\frac{9}{70}+\frac{18}{70}+\frac{2}{70}+\frac{3}{70}=\frac{35}{70}=\frac{1}{2}=0.5$