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A fair coin is flipped twice. Let X be the number of heads in the two tosses, and Y be the indicator r.v. for the tosses landing the same way. That is, Y is equal to 1 if the tosses land in the same way, and 0 otherwise. (a) Find the joint pmf of X and Y.

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Answer:

Explanation:

It can be clearly observed that
X\in{0,1,2} and
Y\in {0,1}

Now lets find the probabilities of all possible combinations


P(X=0, Y=0)=P(Y=0|X=0)P(X=0)=0

Since it is impossible that the coin ha flipped in different ways given that all the time the tail has fallen


P(X=1,Y=0)=P(Y=0|X=1)P(X=1)=1*2*((1)/(2) )^2 =1/2


P(X=2,Y=0)=P(Y=0|X=2)P(X=2)=1*2*((1)/(2) )^2 =1/2

and similarly,


P(X=0,Y=1)=P(Y=1|X=0)P(X=0)=1*((1)/(2) )^2= 1/4


P(X=1,Y=1)=P(Y=1|X=1)P(X=1)=0


P(X=2,Y=1)=P(Y=1|X=2)P(X=2)=1*((1)/(2) )^2=1/4

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