Question

A balanced coin is tossed 3 times, and among the 3 coin tosses, X heads show. Then the same balanced coin is tossed X additional times, and among these X coin tosses, Y heads show.

a. Find the distribution for Y .

b. Find the expected value of Y .

c. Find the variance of Y .

d. Find the standard deviation of Y

Answer #1

a)

below is pmf of Y:

P(Y=0)=P(X=0)*P(Y=0|X=0)+P(X=1)*P(Y=0|X=1)+P(X=2)*P(Y=0|X=2)+P(X=3)*P(Y=0|X=3)

=(_{3}C_{0})*(1/2)^{0}(1/2)^{3}*(_{0}C_{0})*(1/2)^{0}*(1/2)^{0}+(_{3}C_{1})*(1/2)^{1}(1/2)^{2}*(_{1}C_{0})*(1/2)^{0}*(1/2)^{1}+(_{3}C_{2})*(1/2)^{2}(1/2)^{1}*(_{2}C_{0})*(1/2)^{0}*(1/2)^{2}+(_{3}C_{0})*(1/2)^{0}(1/2)^{3}*(_{3}C_{0})*(1/2)^{0}*(1/2)^{3}=27/64

P(Y=1)=P(X=0)*P(Y=1|X=0)+P(X=1)*P(Y=1|X=1)+P(X=2)*P(Y=1|X=2)+P(X=3)*P(Y=1|X=3)

=27/64

P(Y=2)=P(X=0)*P(Y=2|X=0)+P(X=1)*P(Y=2|X=1)+P(X=2)*P(Y=2|X=2)+P(X=3)*P(Y=2|X=3)

=9/64

P(Y=3)=P(X=0)*P(Y=3|X=0)+P(X=1)*P(Y=3|X=1)+P(X=2)*P(Y=3|X=2)+P(X=3)*P(Y=3|X=3)

=1/64

b)

E(Y)=yP(y)=0*(27/64)+1*(27/64)+2*(9/64)+3*(1/64) =0.75

c)

E(Y^{2})=y^2P(y)=0^2*(27/64)+1^2*(27/64)+2^2*(9/64)+3^2*(1/64)
=1.125

Variance =E(Y^{2})-(E(Y))^{2} =0.5625

d)

standard deviation =sqrt(Var(Y))=0.75

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