Problem 3: Each of two people flips a fair coin twice. Let X be the absolute dif

Question

Problem 3: Each of two people flips a fair coin twice. Let X be the absolute difference in the number of heads obtained by two people. Let Y be the sum of number of heads obtained by two people. (a) Construct and write the joint pmf for X, Y. (Use table or formula to write pmf). Make sure to check the requirements for a valid joint pmf. (b) Based on your joint pmf, compute the probability that X2. (c) Based on your joint pmf, compute the probability that X+Y >4. (d) Based on your joint pmf, compute and report the marginal pmf of X. (e) Based on your joint pmf, compute and report the marginal pmf of Y. (f) Explain whether the two variables are independent based on probability.

Explanation / Answer

Sample space for Person 1,2: {HH,HT,TH,TT}

No. of heads obtained for each person can be 0,1,2 each with probability of 0.25, 0.5 and 0.25 repectively

So, X can take value: 0 , 1, 2 (Since X is absolute difference between the no. of heads obtained by Person1 and Person2)

0 (Probability: [0.25*0.25+0.5*0.5+0.25*0.25])

1(Probability:2*[0.5*0.25+0.5*0.25])

2 (Probability:2*[0.25*0.25])

PMF for X:

Y can take values 0,1,2,3,4

0 with probability 0.25*0.25

1 with probability 2*(0.25*0.5)

2 with probability 0.5*0.5+2*(0.25*0.25)

3 with probability 2*(0.25*0.5)

4 with probability 0.25*0.25

PMF for Y:

So joint pmf is

b) Pr. (X<2, Y>2) = Pr. (X=0,1;Y=3,4)=

Pr.(X=0,Y=3)+Pr.(X=0,Y=4)+Pr.(X=1,Y=3)+Pr.(X=1,Y=4) = 0+0.0625+0.25+0=0.3125

c) Pr. (X+Y>4)= Pr.(X=1,Y=4)+Pr.(X=2,Y=3)+Pr.(X=2,Y=4) =0+0+0=0

d) Marginal PMF for X

e) Marginal PMF for Y

f) If two random variables, A and B, are independent, they satisfy the following condition

Pr. (X=0 ? Y=1)=0 but is not equal to Pr.(X=0)*Pr.(Y=1)=0.375*0.25=0.09375

SO X and Y are not independent variable instead they are dependent

X 0 1 2 Probability 0.375 0.5 0.125

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