Answer to Question #225205 in Statistics and Probability for Reyad

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)$\isin$A ], where A is the region {(x,y): x+y<=1}

Answer 1

The total number of refills is: $55+3+3=61$. The probability to get a blue refill is: $p_1=\frac{55}{61}$. For red refills we have: $p_2=\frac{3}{61}$. The joint probability function (probability mass function) $f(x,y)$ is given by: $f(x,y)=p_{X,Y}(x,y)=P(X=x\wedge Y=y)$. Suppose that we selected $2$ refills. There are $m=C_{61}^2=\frac{61!}{2!(61-2)!}=\frac{60\cdot61}{2}=30\cdot61=1830$ ways to do this. Suppose that we selected $2$ refills and among them there are $x\leq2$ blue refills and $y\leq2$ red refills. In addition, since 2 refills are selected. $x+y\leq2$ It is possible to do in $n=C_{55}^xC_{3}^yC_{3}^{2-x-y}=\frac{55!}{x!(55-x)!}\frac{3!}{y!(3-y)!}\frac{3!}{(2-x-y)!(1+x+y)!}$ ways. Thus, we get: $f(x,y)=p_{X,Y}(x,y)=P(X=x\wedge Y=y)=\frac{n}{m}=\frac{C_{55}^xC_{3}^yC_{3}^{2-x-y}}{C_{61}^2}=\frac{\frac{55!}{x!(55-x)!}\frac{3!}{y!(3-y)!}\frac{3!}{(2-x-y)!(1+x+y)!}}{1830}$.

The probability $p[(x,y)\in A]$, where $A=\{(x,y):x+y\leq1\}$ is: $p[(x,y)\in A]=p(X=1,Y=0)+p(X=0,Y=1)=\frac{C_{55}^1C_{3}^0C_{3}^{1}}{C_{61}^2}+\frac{C_{55}^0C_{3}^1C_{3}^{1}}{C_{61}^2}=\frac{55\cdot3}{1830}+\frac{3\cdot3}{1830}=\frac{174}{1830}\approx0.095$