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If I throw 4 dice together, what is the probability that either one of them will show the number 3 ? I tried to calculate it and got to $\frac{(4)}{6}$ (which is highly unlikely to be correct).. any ideas?

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Is this a homework question? If so you should tag it as such. Also, some more detail on how you got $2/3$ would help people answer your question. –  Jeff Jan 9 '12 at 13:24
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"Either" is generally used for a choice between two alternatives, e.g. either the first die shows a $3$ or the second die shows a $3$. Now you have $4$ possibilities to consider. So be sure to distinguish between "at least one of the dice shows a $3$" and "exactly one of the dice shows a $3$". The answers are different in the two cases. –  Dilip Sarwate Jan 9 '12 at 13:32
    
At least one, not exactly one. –  Asaf Jan 9 '12 at 16:59
    
Are these 6 sided dice? I expect that would be a correct assumption, but if they are say 12 sided dice the result is going to be very different. –  aslum Jan 9 '12 at 17:35
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2 Answers

up vote 6 down vote accepted

As mentioned by Dilip Sarwate in the comments, you probably meant to ask either

1) "what is the probability that exactly one of the four shows 3"?

or

2) "what is the probability that at least one of the four shows 3"?

From your proposed answer, I think it would be beneficial to do each of the above problems.

For question 1), finding the probability that exactly one of the dies (dice?) shows three can be computed by taking the ratio of the number of ways this can happen to the number of total possible outcomes:

$$\tag{1} P(\text{exactly one die shows 3})={\text{number of outcomes in which exactly one die shows 3}\over \text{total number of outcomes} }.$$

Now, to compute the required quantities above, it proves convenient to change the problem a bit. Think of throwing one die four times in succession (or think of the dies as having different colors). This will not change the problem.

Let's compute the denominator in (1) first:

There are 6 outcomes for rolling the first die, it can be "1", "2", "3", "4", "5", or "6". Now whatever the first die shows, the second die can show one of 6 numbers. This gives you a total of $6\cdot6=36$ different outcomes for the results of the first two tosses: $$ \matrix{ 11&21&31&41&51&61\cr 12&22&32&42&52&62\cr 13&23&33&43&53&63\cr 14&24&34&44&54&64\cr 15&25&35&45&55&65\cr 16&26&36&46&56&66\cr } $$

For each of these $6^2$ possibilities, the third die can show one of six things. This gives a total number of possibilities of $6^2\cdot 6=6^3$ outcomes for rolling three times.

It should be apparent that the number of possible outcomes from rolling four times is $6^4$. So the denominator in (1) is $6^4$.

What I used above is the multiplication principle for counting the number of outcomes of a sequence of experiments.

Now, on to finding the numerator in (1): the number of outcomes in which exactly one die shows 3.

We can break this down into cases: either only the first roll was 3, or only the second roll was 3, or only the third roll was 3, or only the forth roll was 3.

These cases are mutually exclusive: no two of them can occur at the same time. Moreover, if exactly one roll was 3, then one of these four events occur.

To find the total number of outcomes when you break things up into mutually exclusive and exhaustive cases as above, you figure out the size of each case and then add them.

So, let's do that. The total number of ways case 1 can occur is, using the multiplication principle: $1\cdot5\cdot5\cdot5$ (there is one outcome for the first toss since we know it is a 3; there are 5 outcomes for the second toss since we know it is not a 3, etc..).

The other three cases can be handled in a similar manner. Each turns out to be $5^3$. So the number of outcomes for which exactly one die shows 3 is $5^3+5^3+5^3+5^3=4\cdot 5^3$.

Thus, the numerator in (1) is $4\cdot 5^3$; and thus, recalling that the denominator in (1) is $6^4$, the probability that exactly one die shows 3 is $4\cdot 5^3\over 6^4 $.

For question 2), you should follow Rasmus' advice and calculate the probability that it is not the case that at least one die shows 3. This would be the probability that no die shows 3; which you can readily calculate using the multiplication principle and the formula $$ P(\text{no die shows 3})={\text{number of outcomes in which no die shows 3}\over \text{total number of outcomes} }.$$

Well, that's the probability that no die shows 3. Call this number $x$. No die shows three $x\cdot 100$ percent of the time; so what percentage of the time does the complement occur? Well, it's $100-x\cdot100$...

So, the probability that at least one die shows 3 is $1-P(\text{ no die shows 3})$.

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[Edit: I at first didn't notice the ambiguity in the question. The following hint is only helpful if you meant to ask for the probability of at least one of the dice showing a $3$.]

Hint: Try to compute the probability of the converse. What is the probability of one specific dice not showing the number $3$? What is then the probability of all four dice not showing the number $3$?

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