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An urn contains 11 balls, 3 white, 3 red, and 5 blue balls. Take out two balls at random, without replacement. You win $1 for each red ball you select and lose a $1 for each white ball you select. Let X be the random variable that notes the amount you win. Find the probability mass function (p.m.f.) of X.

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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
X=-2):^3C_2=3


P(X=-2)=(3)/(55)

The number of ways so that two balls are red (and
X=2):^3C_2=3


P(X=2)=(3)/(55)

The number of ways so that one ball is red, one is white (and
X=0):^3C_1*^3C_1=9

The number of ways so that two balls are blue (and
X=0 ):
^5C_2=(5 \cdot 4) / 2=10

i.e.
P(X=0)=(10+9)/(55)=(19)/(55)

The number of ways so that one ball is blue, one is white (and
X=-1 ):
^5C_1*^3C_1=15


P(X=-1)=(15)/(55) =(3)/(11)

The number of ways so that one ball is blue, one is red (and
X=1 ):
^5C_1*^3C_1=15


P(X=1)=(15)/(55) =(3)/(11)

Thus, the probability mass function (p.m.f.) of X would be ( in attachment) :

An urn contains 11 balls, 3 white, 3 red, and 5 blue balls. Take out two balls at-example-1
User Yerlan
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