1
$\begingroup$

Consider an experiment of tossing two coins, denoted by $a$ and $b$, three times. Coin $a$ is fair (i.e., the probability of the head or the tail is $1/2$), while the coin $b$ is not and its probability of the head is $1/4$ and the tail is $3/4$. Let $X$ be the number of heads resulting in the three tosses of the coin $a$, and let $Y$ be the number of heads resulting in the three tosses of the coin $b$. Consider bivariate r.v. $(X; Y )$.

1 -Find the pmf of $(X,Y)$.

Progress: I have found the probabilities of $X$ and $Y$ separately meaning I have $P(x=0)$,$P(x=1)$, $P(x=2)$, $P(x=3)$ and similarly for $Y$. Now I don't know how to put them all together.

2- Find $P(X > Y)$

1 Answers 1

2

With $\mathbb{P}[X=i]$ and $\mathbb{P}[Y=i]$ figured out for $i\in\{0,1,2,3\}$ you are done, since $\mathbb{P}[X=i,Y=j]=\mathbb{P}[X=i]\mathbb{P}[Y=j]$ for any $i\in\{0,1,2,3\}$ and $j\in\{0,1,2,3\}$ since the two coin tosses are independent.

$\mathbb{P}[X>Y]$ is just sum of $\mathbb{P}[X=i,Y=j]$ over $i\in\{0,1,2,3\}$ and $j\in\{0,1,2,3\}$ such that $i>j$.

  • 0
    Could you elaborate on the pmf? I usually see them as a table or function with multiple lines (forgot the actual math term)?. for P(X>Y), I don't understand what you mean by summing them with i>J. Does that mean it is P(x=0 to 3) * P(Y=0 to 2)? – 2017-01-23
  • 0
    @user3064033: You are after a 4 by 4 table where entry in row i column j is P[X=i,Y=j]. My answer allows you to calculate P[X=i,Y=j]. And P[X>Y]=P[X=1,Y=0]+P[X=2,Y=0]+P[X=2,Y=1]+P[X=3,Y=0]+P[X=3,Y=1]+P[X=3,Y=2]. – 2017-01-23