# a coin toss and a die roll simultaneously

Suppose that we toss a fair coin and, then toss a fair four-sided die.

(a) Defne the sample space for this experiment.

(b) Let $$X$$ be the random variable defined as (number of heads) + (die score modulus 2). Identify the possible values $$x$$ that $$X$$ can have, and write down the event $$X = x$$ for each value of $$x$$. Then calculate the probability mass function for $$X$$.

Solution: (a) For the coin toss, there are two possible outcomes head(H) and tail(T).

For the four-sided die roll there are four possible outcomes $$1,2,3,4$$.

So, the sample space is $$\{(H,1),(H,2),(H,3),(H,4),(T,1),(T,2),(T,3),(T,4)\}$$.

I am unable to understand the problem (b). It says that $$X$$ be the random variable defined as (number of heads) + (die score modulus 2). But there are only one head with die score $$2$$. that is $$(H,2)$$. Is it ? I think I am not correct. What does it mean actually? and how to solve this?

• You can have either $0$ or $1$ heads from the coin flip and the four-sided die can give you either $0$ or $1 \pmod 2$. Commented Mar 31, 2023 at 8:14
• Example: you toss the coin and roll the die. The coin comes up tails. The die comes up 3. Then (number of heads) = 0, and (die score modulus 2) = 1. Thus the value of $X$ is $0 + 1 = 1$. You can do this for all coin and die outcomes. ("Die score modulus 2" means "the remainder when dividing the die score by 2", so a die score of 1 gives you 1, a die score of 2 gives you 0, a die score of 3 gives you 1, a die score of 4 gives you 0. Normally we would use the world "modulo" here, not "modulus".) Commented Mar 31, 2023 at 8:25

Indeed the sample space is $$\Omega=\{H,T\}\times\{1,2,3,4\}$$.
For (b) it is simpler to write $$X=Y+Z$$ where $$Y$$ is the number of heads and $$Z$$ the die score mod $$2$$. Then $$\mathbb P(Y=0) = \frac12=\mathbb P(Y=1)$$ and $$\mathbb P(Z=0)=\frac12=\mathbb P(Z=1)$$ (note that $$Z=0$$ when the die rolls $$2$$ or $$4$$ and $$Z=1$$ when the die rolls $$1$$ or $$3$$). By independence of $$Y$$ and $$Z$$ we have $$\mathbb P(X=k) = \begin{cases} \mathbb P(Y=0)\mathbb P(Z=0) = \frac14,& k=0\\ \mathbb P(Y=0)\mathbb P(Z=1)+ \mathbb P(Y=1)\mathbb P(Z=0) = \frac12, & k=1\\ \mathbb P(Y=1)\mathbb P(Z=1) = \frac14,& k=2. \end{cases}$$