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

Three dice are thrown once and the random variable $X$ denotes the number of dice that brought an even number. Find:

  1. The probability function of $X$.
  2. The cumulative distribution function of $X$.
share|cite|improve this question
Think of it as a Bernoulli trial. – hardmath Oct 31 '11 at 0:39
Proceeding from @hardmath's comment, $X$ must have a binomial distribution $\mathrm{Bin}(3, 0.5)$. Can you see why? – Srivatsan Oct 31 '11 at 2:00
up vote 2 down vote accepted

The probability that a die shows an even number is $1/2$, so the problem is essentially the same as the problem of the probabilities of $0$ heads, $1$ head, $2$ heads, $3$ heads when we toss three fair coins. We get $$f_X(0)=\frac{1}{8}; \qquad f_X(1)=\frac{3}{8}; \qquad f_X(2)=\frac{3}{8}; \qquad f_X(3)=\frac{1}{8}.$$

For the cumulative distribution function $F_X(x)$, recall that $F_X(x)=P(X\le x)$, and remember that $F_X(x)$ is defined for all real numbers $x$. In general, $F_X(x)$ is the "weight" up to and including $x$.

Then we can see that $F_X(x)=0$ for all $x<0$. For $0 \le x <1$, we have $F_X(x)=\frac{1}{8}$. For $1 \le x <2$, we have $F_X(x)=\frac{4}{8}$. For $2 \le x <3$, we have $F_X(x)=\frac{7}{8}$. And finally, for $x \ge 1$, we have $F_X(x)=1$. The cumulative distribution function jumps upwards at $x=0$, $1$, $2$, and $3$.

share|cite|improve this answer
thanks for your help – kary Nov 2 '11 at 11:20

It's the probability distribution of the number of successes in three independent trials with probability $1/2$ of success on each trial, so it's a binomial distribution.

share|cite|improve this answer
+1 for the words "binomial distribution" ;) – Srivatsan Oct 31 '11 at 2:56
thanks for your help – kary Nov 2 '11 at 11:15

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.