Tara Alvarado

Answered

2021-12-18

Two fair dice are rolled. Find the joint probability mass function of X and Y when (a) X is the largest value obtained on any die and Y is the sum of the values; (b) X is the value on the first die and Y is the larger of the two values; (c) X is the smallest and Y is the largest value obtained on the dice.

Answer & Explanation

Stella Calderon

Expert

2021-12-19Added 35 answers

Define $N}_{1$ and $N}_{2$ as random variables that mark numbers obtained on the first and the second die. We know that $N}_{1$ and $N}_{2$ are independent and that $N}_{1},{N}_{2$

a) Here we have that $X=max({N}_{1},{N}_{2})$ and $Y={N}_{1}+{N}_{2}$. Thus $X\in \{1,\dots ,6\}$ and $Y\in \{2,\dots ,12\}$. Also we have that X<Y almost certainly. So take any $k<l$, where k and l are from the ranges given above. Consider event $X=k,Y=l$. That menas that the maximum value on any die is k and that the sum on both dice is l. Observe that the only possible pairs of $({N}_{1},{N}_{2})$ corresponding to that event are $(k,l,-k)$ and $(l,-k,k)$ if $l<2k$. If $l=2k$, the only possible pair s (k,k). Hence the required PMF is

$P(X=k,Y=l)=\{\begin{array}{ll}\frac{2}{36}& ,k<l<2k\\ \frac{1}{36}& ,l=2k\end{array}$

rodclassique4r

Expert

2021-12-20Added 37 answers

Here we have that $X={N}_{1}$ and $Y=max({N}_{1},{N}_{2})$ . Observe that both variables are in $\{1,\dots ,6\}$ and that $X\le Y$ almost certainly. Take any $k\le l$ from the range given above. We have that

$P(X=k,\text{}Y=l)=P(Y=l\mid X=k)P(X=k)$

Suppose that$k=l$ and that we are given $X=k$ . In that case, $N}_{2$ can be any number from the range $1,\dots ,k$ to obtain the required $Y=l$ . Hence

$P(Y=l\mid X=k)P(X=k)=\frac{k}{6}\cdot \frac{16}{=}\frac{k}{36}$

If$k<l$ and we are giveen that $X=k,{N}_{2}$ must be equal to l obtain $Y=l$ . So, in that case

$P(Y=l\mid X=k)P(X=k)=\frac{16}{\cdot}\frac{16}{=}\frac{1}{36}$

Suppose that

If

RizerMix

Expert

2021-12-29Added 437 answers

Here we have that

Suppose that k<l. In this case we have to have

if k=l, the only possibility is

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