Answer with explanation:
Given: An urn contains 11 balls, 3 white, 3 red, and 5 blue balls.
You win $1 for each red ball you select and lose a $1 for each white ball you select.
Let X be the number of times you win.
The total number of ways to select 2 balls (order does not matter) =
The number of ways so that two balls are white (and [tex] X=-2):^3C_2=3 [/tex]
[tex]P(X=-2)=\dfrac{3}{55}[/tex]
The number of ways so that two balls are red (and [tex] X=2):^3C_2=3 [/tex]
[tex]P(X=2)=\dfrac{3}{55}[/tex]
The number of ways so that one ball is red, one is white (and [tex] X=0):^3C_1\times^3C_1=9 [/tex]
The number of ways so that two balls are blue (and [tex] X=0 [/tex] ): [tex] ^5C_2=(5 \cdot 4) / 2=10 [/tex]
i.e. [tex]P(X=0)=\dfrac{10+9}{55}=\dfrac{19}{55}[/tex]
The number of ways so that one ball is blue, one is white (and [tex] X=-1 [/tex] ): [tex] ^5C_1\times^3C_1=15 [/tex]
[tex]P(X=-1)=\dfrac{15}{55} =\dfrac{3}{11}[/tex]
The number of ways so that one ball is blue, one is red (and [tex] X=1 [/tex] ): [tex] ^5C_1\times^3C_1=15 [/tex]
[tex]P(X=1)=\dfrac{15}{55} =\dfrac{3}{11}[/tex]
Thus, the probability mass function (p.m.f.) of X would be ( in attachment) :
