Jack and Jill are independently performing independent Bernoulli trials. For con

Question

Jack and Jill are independently performing independent Bernoulli trials. For concreteness, assume Jack tosses a coin with probability of heads 0.85 and Jill tosses a coin with probability of heads 0.45. Let X1,X2,… be Jack's results and Y1,Y2,… be Jill's results, with XiBern(0.85) and YiBern(0.45) .

A. Find the the expected number of tosses until the first time they have simultaneous heads, i.e. the smallest n for which Xn=Yn=1 . (Hint: Define a new sequence of independent Bernoulli trials Zi=XiYi and use the story of the Geom/FS distribution.)

Answer:

B. Find the the expected number of tosses until the first time at least one of them tosses a head. (Hint: Define a new sequence of Bernoulli trials and use the story of the Geom/FS distribution.)

Answer:

C. Find the the expected number of tosses until the 17th time they have simultaneous heads. Answer:

Explanation / Answer

P(X=0,Y=0)=P(X=0).P(Y=0)=0.15*0.55=0.0825

P(X=0,Y=1)=P(X=0).P(Y=1)=0.15*0.45=0.0675

P(X=1,Y=0)=P(X=1).P(Y=0)=0.85*0.55=0.4675

P(X=1,Y=1)=P(X=1).P(Y=1)=0.85*0.45=0.3825

Let Z=XY

(A)  Now Consider the random variable Z to follow geometric distribution ie success is occured if both jack and jill got heads at the same time and failure if not

P(Z=0)=P(X=0,Y=0)+P(X=0,Y=1)+P(X=1,Y=0)=0.0825+0.0675+0.4675=0.6175

P(Z=1)=P(X=1,Y=1)=1-0.6175=0.3825

Expected value in the case of geometric distribution is q/p=0.6175/0.3825=1.6~2

The success is expected in the 2nd toss

B) Again Z is considered to be a geometric random varible with success slightly redefined that is atleast one of them tosses a head

P(Z=0)=q=0.0825

P(Z=1)=p=0.9175

Expected value in this case is q/p=0.09

The success is expected in the very first toss

C) Here the random variable we consder Zi  to follow geometric distribution ie success is occured if both jack and jill got heads at the same time and failure if not   and we define W=Z1+Z2+....+Z17 which follows negative binomial distribution

E(W) =17*q/p=17*1.6=27.2

Hence we expect the heads to be together for the 17th time in the 27th toss

P(X=x,Y=y) y 0 1 Total x 0 0.0825 0.0675 0.15 1 0.4675 0.3825 0.85 0.55 0.45 1

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