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Questions: You roll a fair die 18 times; the rolls are independent of each other. What is the probability that you roll a 5 exactly three times?

Answer: 0.245198448

Attempt: From the basics, I have learned so far, I need to determine the sample space and the event for this case and then do event/sample space

Sample Space: {1,2,3,4,5,6}*18 = 1/6*18=3 Event: Roll a 5 exactly 3 times, so 1/6*3=1/2

Pr(A) = 1/2 / 3 = 0.1667

I'm not confident with how I determined my sample space or event. Any help on how to approach these steps logically would help.

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This sounds like the kind of thing you would, ideally, use a binomial distribution for, if you're aware of that sort of thing.

For the record, your calculations are completely wrong.

For example, the sample space consists of all possible rolls you could make - 6 different outcomes for each of the 18 independent rolls. This completely sets you on the wrong track from the get-go. For example, one event in the space would consist of the rolls

$$(1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3)$$

Notice how the first value has 6 possible values, as does the second, third, fourth, etc., all of the way up to the 18th value. That's why you have $6^{18}$ possible events in your sample space.

You also need to find how many possible events in the sample space have three rolls of 5. This is why I specified the binomial distribution earlier, because it makes things like these much, much simpler to look at. For example, if you want the probability of $r$ events occurring in $n$ trials, you would find that

$$P(\text{r occurrences in n trials}) = \binom{n}{r}P(\text{occurrence})^r (1 - P(\text{occurrence}))^{n-r}$$

That gives you the possible number of desired events in the sample space when $n = 18, r = 3$. (The probability of rolling a 5 is $1/6$, obviously.)

The 3 rolls of 5 could occur in various places in the 18 rolls (they could be the first three rolls, last three, spread out all over, etc.) so we use the binomial coefficient $\binom{n}{r}$ to account for the variations in ordering. This method also has the benefit of giving you the probability outright since it accounts for the whole "number of desired events divided by the number of events in the sample space" phenomenon.


For the record this is equivalent to Siong Thye Goh's calculation in the other answer that's up, as

$$\frac{1^3 \cdot 5^{15}}{6^{18}} = \left( \frac{1}{6} \right)^3 \cdot \left( \frac{5}{6} \right)^{15}$$

which gives you the probabilities in the formula I noted earlier. I personally prefer the definition in terms of the probability as I gave it but it's a matter of opinion I guess.

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  • $\begingroup$ This is the first time I've seen this formula, how often can I apply this formula for these types of problems? Any restrictions? $\endgroup$
    – Toby
    Nov 26 '18 at 1:11
  • $\begingroup$ Basically to any problem that asks for "what is the probability of (doing this thing) exactly $r$ times out of a total of $n$ trials?" If you want to find it in "at most" or "at least" $r$ tries, you could use it, just summing over the appropriate values of $r$. An important nuance is that each trial/event has to be independent from the previous one as well for the formula to work. $\endgroup$ Nov 26 '18 at 1:52
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The sample space are $\{1,\ldots,6\}^{18}$, we are interested in those of which exactly $3$ of them takes value $5$. Out of the $18$ positions, we have to pick $3$ of them to take value $5$.

$$\frac{\binom{18}{3}\cdot 1^3\cdot 5^{15}}{6^{18}}$$

Relevant keyword: Binomial distribution.

Remark about your attempt:

$\{1,\ldots,6\}^{18}$ has cardinality $6^{18}$ rather than $6(18)$.

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  • $\begingroup$ where is the 1^3 coming from? $\endgroup$
    – Toby
    Nov 26 '18 at 0:59
  • $\begingroup$ roll a $5$ exactly $3$ times. $\endgroup$ Nov 26 '18 at 1:01
  • $\begingroup$ So, shouldn't it be 5^3?, I'm a little confused $\endgroup$
    – Toby
    Nov 26 '18 at 1:02
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    $\begingroup$ $5$ is just a label. For those special $3$ locations where we want it to be $5$, there is only one possibility. For the other $15$ locations, there are $5$ options. $\endgroup$ Nov 26 '18 at 1:03

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