An urn contains 1 black ball and 2 white balls. One ba is drawn at random and it

Question

An urn contains 1 black ball and 2 white balls. One ba is drawn at random and its color is noted. The ball is replaced in the urn, together with an additional ball of its color so there are now four balls in the un Again, one ball is drawn at random from the urn, then replaced along with an additional ba of its color The process continues this way. (a) Let Bn denote the number of black balls in the urn Just before the nth ball is drawn (ie. Bi = 1). For n 2 1, find E(Bn+Bn (b) Find E(B2) and E(B)

Explanation / Answer

the urn contains 1 black ball and 2 white balls.

one ball id drawn at random and its color is noted. The ball is replaced in the urn, together witha na additional ball of its color. so number of balls increased by 1.

and this process continues the same way.

a) Bn denote the number of black balls in theurn just before the nth ball is drawn

so B1=1

B2 is either 1 or 2. and total number balls in the urn is 4

so for Bn the number of balls in the urn will be n+2

we need to find E[Bn+1|Bn]

let Bn=x is given

so out of n+2 balls there are x black balls and n+2-x white balls.

so Bn+1 can either be x or x+1

now P[Bn+1=x|Bn=x]=P[drawing a white ball]=(n+2-x)/(n+2)

and P[Bn+1=x+1|Bn=x]=P[drawing a black ball]=x/(n+2)

so E[Bn+1|Bn]=x*(n+2-x)/(n+2)+(x+1)*x/(n+2)=(nx+2x-x2+x2+x)/(n+2)=(nx+3x)/(n+2) [answer]

b) E[B2] and E[B3] needs to be calculated

now B2 can either be 1 or 2

so P[B2=1]=P[a white ball is drawn initially]=2/3

P[B2=2]=P[a black ball was drawn initially]=1/3

so E[B2]=1*2/3+2*1/3=4/3

now we need to find E[B3]

now E[B3]=E[E[B3|B2]]

now B3|B2=b can take values b or b+1

now P[B3=b|B2=b]=P[a white bal is drawn from 4 balls with b black balls and 4-b white balls]=(4-b)/4

P[B3=b+1|B2=b]=P[a white bal is drawn from 4 balls with b black balls and 4-b white balls]=b/4

so E[B3|B2]=b*(4-b)/4+(b+1)*b/4=(4b-b2+b2+b)/4=5b/4

hence E[B3]=E[E[B3|B2]]=E[5B2/4]=5*E[B2]/4=5*4/(4*3)=5/3 [answer]

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