Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can some one please tell me how to do it?

Let $(i,j)$ denote the numbers on the top faces when a pair of fair dice is rolled.

Let $X(i,j)=i+j-3$.

Find the range of $X$.

Find the probability distribution function $f_{X}(x)$ of $X$.

Find each of the following

  1. $p(x>11)$
  2. $p(x<7)$
  3. $p(x \ge0)$
share|cite|improve this question
Where are you stuck? What have you tried? – Inquest May 23 '12 at 11:03
up vote 2 down vote accepted

The range: The values of $i$ and $j$ range from $1$ to $6$. So the smallest possible value of $i+j$ is $2$, and the largest is $12$. So $i+j-3$ ranges over the integers from $-1$ to $9$.

Thus the range of $X$ is the set of all integers $n$ such that $-1\le n\le 9$. You could write it also as $\{-1,0,1,1,3,4,5,6,7,8,9,\}$.

The probability distribution function: Recall that $f_X(x)=P(X=x)$. We need to specify $f_X(x)$ for all $x$ in the range.

As a start, we find $f_X(-1)=P(X=-1)$. We have $X=-1$ precisely if $i+j=2$, which happens if both dice show a $1$. The probability of this is $\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{36}$.

Next we find $f_X(0)$. We have $X=0$ precisely if $i+j=3$. This happens if the yellow die shows $1$ and the pink shows $2$, or the other way around. So $f_X(0)=\frac{2}{36}$.

Next we find $f_X(1)$. Note that $X=1$ precisely if $i+j=4$, that is, if two dice give sum $4$. We have probably long known that the sum of two fair dice is $4$ with probability $\frac{3}{36}$.

Let's get slightly more abstract. Let $Y=X+3$. Then $Y=i+j$ is just the sum of two dice, and $X=k$ iff $Y=X+k$. So $f_X(1)=f_Y(4)=\frac{3}{36}$.

Similarly, $f_X(2)=f_Y(5)=\frac{4}{36}$, $f_X(3)=f_Y(6)=\frac{5}{36}$, $f_X(4)=f_Y(7)=\frac{6}{36}$, $f_X(5)=f_Y(8)=\frac{5}{36}$, and so on.

The numbered questions: $1$. We want $P(X\gt 11)$. Since $X=Y-3$, We have $X\gt 11$ precisely if $Y \gt 14$. Of course the sum of two dice cannot be $\gt 14$, and therefore $P(X\gt 11)=0$.

$2$. This one is more work. The most mechanical way to do it is to observe that $X \lt 7$ precisely if $X$ takes on the values $-1$ to $6$. So add up the values of $f_X(k)$ obtained earlier, $k=-1$ to $k=6$.

Or else note that this is the probability that $Y \lt 10$, that is, the probability that the sum $Y$ of two dice is less than $10$. It is easier to find the probability that $Y\ge 10$. This is $\frac{3}{36}+\frac{2}{36}+\frac{1}{36}=\frac{6}{36}$. So the probability that $Y \lt 10$ is $\frac{30}{36}$.

$3$. More of the same.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.