Question

What is the probability of rolling a total of $7$ with two dice at least once in $10$ rolls?

Answer 1

Hint: Probability is a branch of mathematics concerned with empirical representations of the likelihood of an occurrence occurring or the truth of a proposition. The probability of an event is a number between $0$ and $1$, with $0$ indicating impossibility and $1$ indicating certainty.

Complete step-by-step solution:
A six-sided cube with the numbers $1-6$ on the faces is the most common type of die.
When we roll a die there are $6$possible outcomes. Hence, there are $36$ possible outcomes when flipping two dice.
Six of the $36$ outcomes have a total of seven:
$\left\{1+6,2+5,3+4,4+3,5+2,6+1\right\}$
That is, $30$ of the $36$ possible outcomes will not be a total of $7$.
${\displaystyle \frac{30}{36}}={\displaystyle \frac{5}{6}}$
${\displaystyle \frac{5}{6}}$ of the time, we will not get a total of $7$ on the first roll.
i.e., We didn't get a $7$ on the first roll ${\displaystyle \frac{5}{6}}$ of the time.
${\displaystyle \frac{5}{6}}$ of the time, we will not get $7$ on the second roll.
That is, we will not get a total of $7$ on either of the first two rolls ${\displaystyle \frac{5}{6}}\times {\displaystyle \frac{5}{6}}={\left({\displaystyle \frac{5}{6}}\right)}^2$ of the time.
Following this logic, we can see that ${\left({\displaystyle \frac{5}{6}}\right)}^{10}$ of the time, we will not get a total of $7$ on any of the first $10$rolls.
i.e.; Approximately $16.5\mathrm{\%}$ of the time, we will not get a total of $7$on any of the first $10$ rolls.
This means that at least one of the first ten rolls would result in a total of $7$.
$100\mathrm{\%}-16.5\mathrm{\%}=83.85\mathrm{\%}$
Hence, the probability of rolling a total of $7$ with two dice at least once in $10$ rolls is $83.85\mathrm{\%}$

Note: The higher an event's probability, the more likely it is that it will occur. The underlying dynamics and regularities of complex systems are often defined using probability theory.