Answer to Question #225531 in Statistics and Probability for Riyad

Question

Two refills for a ball point pen are selected at random for a box that contains 55

blue, 3 red and 3 green refills. If X is the number of blue refills and Y is the number

of red refills selected .Find the joint probability function f (x, y) and p[(x,y)∈A ], where A is the region {(x,y): x+y<=1}

Answer 1

Part A

The possible pairs of values (x, y) are (0, 0), (0, 1), (1, 0), (1, 1), (0, 2), and (2, 0), where p(0, 1), for

instance stands for the probability that a red and a green refill are selected. The total number of equally likely ways of selecting any 2 refills from the 8 is:

$\begin{pmatrix} 61 \\ 2 \end{pmatrix}= \frac{61!}{2!58!}=107970$

The number of ways of selecting 1 red from 3 red refills and 1 green from 3 green refills is

$\begin{pmatrix} 3 \\ 1 \end{pmatrix}\begin{pmatrix} 3 \\ 1 \end{pmatrix}=9$

Hence, $p(1, 1) = \frac{9}{107970}$ .

Similar computations provide the probability for the other instances, which are shown in the table below. It is worth noting that the probability add up to one.

$p(x,y)= \frac{\begin{pmatrix} 55 \\ x \end{pmatrix}\begin{pmatrix} 3 \\ y \end{pmatrix}\begin{pmatrix} 3 \\ 3-x-y \end{pmatrix}}{\begin{pmatrix} 61 \\ 2 \end{pmatrix}}\\ For \space x =0,1,2\\ For \space y =0,1,2\\ 0≤x+y≤2$

Part B

$P[(X, Y) \in A] = P(X + Y≤1)\\ = p(0, 0) + p(0, 1) + p(1, 0)\\ =\frac{55}{107970}+\frac{3}{53985}+\frac{165}{107970}\\ =\frac{113}{53985}$