Arthur Pratt

2022-01-05

If a die is rolled 30 times, there are ${6}^{30}$ different sequences possible. The following question asks how many of these sequences satisfy certain conditions.
What fraction of these sequences have exactly three 4s and three 5s?

### Answer & Explanation

Laura Worden

Let 'S' be the sample space of rolling dice once.
$S=\left\{1,2,3,4,5,6\right\}$
Assuming a dice is fair, all the outcomes are equally likely.
When a dice is rolled once, there are $6$ possible outcomes, when a dice is rolled twice, there are ${6}^{2}$ possible outcomes, and, when the dice is rolled $n$ times, there are ${6}^{n}$ possible outcomes.
Here, the dice is rolled $30$ times. Thus, the total number of possible outcomes are ${6}^{30}$.
There are $30$ places available in each sequence, the number of ways of selecting exactly three $4S=30{C}^{3}$
In the remaining $27$ places, the number of ways of selecting exactly three $5S=27{C}^{3}$
In the remaining $24$ places, any number other than $4$ and $5$ can take place.
The number of possible ways of filling the remaining $24$ places $={4}^{24}$
The total number of sequences with exactly three $4s$ and three $5s$ $=\left(30{C}^{3}\right)\left(27{C}^{3}\right)\left({4}^{24}\right)$
Use the formula to compute the required probability:

$=\frac{\left(4060\right)\left(2925\right)\left(281474976710656\right)}{221073919720733000000000}=0.01512$

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