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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!

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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.

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  • $\begingroup$ 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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