Question

- Three balls are randomly chosen from an urn containing 3 white, 4 red and 5 black balls. Suppose one will win $1 for each white ball selected, lose 1$ for each red ball selected and receive nothing for each black ball selected. Let Random Variable X denote the total winnings from the experiment. Find E(X).

Answer #1

below is pmf of X:

P(X=-3)=P(all three are red
balls)=^{4}C_{3}/^{12}C_{3}
=1/55

P(X=-2)=P(2 red and 1 black
balls)=^{4}C_{2}*^{5}C_{1}/^{12}C_{3}
=3/22

P(X=-1)=P(2 red and 1 white balls)+P(1 red and 2 black
balls)=^{3}C_{1}^{4}C_{2}*^{5}C_{0}/^{12}C_{3}
+^{4}C_{1}*^{5}C_{2}/^{12}C_{3}
=29/110

P(X=0)=P(all three are black)+P(1 red, 1black and 1 white
ball)=^{5}C_{3}/^{12}C_{1}
+^{3}C_{1}^{4}C_{1}*^{5}C_{1}/^{12}C_{3}
=7/22

P(X=1)=P(1 red and 2 white balls)+P(1 white and 2 black balls)=21/110

P(X=2)=P(2 white and 1 black balls)=3/44

P(X=3)=P(all three are white balls)=1/220

hence E(X)=xP(x)=-3*(1/55)+(-2)*(3/22)+(-1)*(29/110)+0*(7/22)+1*(21/110)+2*(3/44)+3*(1/220)=
-1/4 =**$ -0.25**

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