 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? 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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