joint probability of dice and coin

Question

I have a question where $X_1$ is the number on a die $(1,2,3,4,5,6)$ and $X_2$ is the number of die plus the value of indicator on coin $(H=1, T=0)$. Find the joint probability of $X_1$ & $X_2$ , when die and coin tossed together. How will we find the joint distribution ? Thanks for positive response

Answer 1

Let $C$ denote the number of the coin

In this situation most likely you are allowed to assume that $X_1$ and $C$ are independent random variables.

Moreover the random variables are discrete and then the joint PMF can be found as:$$P((X_1,C)=(x_1,c))=P(X_1=x_1,C=c)=P(X_1=x_1)P(C=c)$$where the second equality is a consequence of the independence.

Characteristic for independent random variables: if the marginal distributions are known then also the joint distribution can be found. This is not so if independence lacks.

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Now observe that: $\{X_1=x_1,X_2=x_2\}=\{X_1=x_1,C=x_2-x_1\}$ so that $$P(X_1=x_1,X_2=x_2)=P(X_1=x_1,C=x_2-x_1)=P(X_1=x_1)P(C=x_2-x_1)$$

I leave it up to you to find for what pairs $(x_1,x_2)$ this probability is positive.

Answer 2

Only the value pairs $(n,n)$ and $(n,n+1)$, where $n=1,\ldots 6$ are possible, so 12 possible pairs.

Then $P(X_1 = n \land X_2 = n) = \frac{1}{6}\cdot \frac{1}{2}= \frac{1}{12}$ because we have to throw $n$ with the die and tails with the coin, independently.

Similarly $P(X_1 = n \land X_2 = n+1) = \frac{1}{6}\cdot \frac{1}{2}= \frac{1}{12}$ too. All $12$ possible outcome pairs are equiprobable, with fair coin and die.