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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 $X$.
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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

2 Answers 2

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 $3$.

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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.

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

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