A game consists of first rolling an ordinary 6-sided die once and then tossing a fair coin once. The score, which consist of adding the number of spots showing on the die to the number of heads showing on the coin, is a random variable called X.

a) Give the probability function for this random variable b) Give the CDF for this random variable c) Find P(X>3) d) Find the probability that the score is an odd integer

I'm confused what my supp(X) is? If anyone can help with this problem that would be awesome, thanks!


In the discrete case, the support of the random variable $X$ is the set of values of $x$ such that $\Pr(X=x)\ne 0$. In our case, the possible values of $X$ range from $1$ to $7$, so the support of $X$ is $\{1,2,3,4,5,6,7\}$.

Added: We find the distribution of $X$, by specifying $\Pr(X=x)$ for all values $x$ in the support of $X$.

In order for $X$ to be $1$, we need to roll a $1$ and toss a tail. The probability of this is $\frac {1}{6}\cdot \frac{1}{2}$. Thus $\Pr(X=1)=\frac {1}{12}$.

The random variable $X$ can be $2$ in two ways: (i) we get a $2$ on the die, and roll a tail or (ii) we roll a $1$ on the die, and toss a head. The probability of (i) is $\frac{1}{6}\cdot\frac{1}{2}$. The probability of (ii) is the same. It follows that $\Pr(X=2)=\frac{1}{6}$.

You can handle the probabilities that $X=3$, $X=4$, and so on to $7$.

For the cdf $F_X(x)$, recall that $F_X(x)$ is the probability that $X\le x$, and is defined for all real $x$.

If $x\lt 1$, the $\Pr(X\le x)=0$, so $F_X(x)=0$.

If $1\le x\lt 2$, then $\Pr(X\le x)=\frac{1}{12}$, so in this interval $F_X(x)=\frac{1}{12}$.

If $2\le x\lt 3$, then $\Pr(X\le x)=\frac{1}{12}+\frac{1}{6}$. Thus $F_X(x)=\frac{3}{12}$ in this interval.

Continue. Don't forget about $F_X(x)=1$ if $x\ge 7$.

The remaining questions will probably not cause any difficulty.

  • $\begingroup$ why is my support of X going to 7, when a die has a 6 sides? I am confused about that.. $\endgroup$ – Nikola Sep 17 '13 at 14:28
  • $\begingroup$ I got it, Thank you so much for help! $\endgroup$ – Nikola Sep 17 '13 at 14:41
  • $\begingroup$ You are welcome. You got it, the support (for discrete) consists of all values the rv takes on with non-zero probability. The numbers $1$ throgh $7$ can all be the values of $X$. $\endgroup$ – André Nicolas Sep 17 '13 at 15:42

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