Question

On a roll Suppose that you roll a fair, six-sided die 10 times. What’s the probability that you get at least one 6?

Short answer

The probability of getting at least one $6$ is$0.8385.$

Step-by-step solution

Given information

Fair, six-sided die-rolled$10$times.
Such that
At least 1 of the 10 rolls results in a $6$

Calculation

Two events are independent if the probability of one event's occurrence has no bearing on the probability of the other event's occurrence.

According to the multiplication rule for independent events,

$P(\mathrm{A}\text{and}\mathrm{B})=P(\mathrm{A}\cap \mathrm{B})=P(\mathrm{A})\times P(\mathrm{B})$

According to the complement rule,

$P\left({\mathrm{A}}^{\mathrm{c}}\right)=P\left(\text{not}\mathrm{A}\right)=1-P(\mathrm{A})$

Let,

$A$: One roll results in a $6$

$B$: At least 1 of the 10 rolls results in a $6$

$A^c:$One roll does not result in a $6$

$B^c$: None of the 10 rolls results in a $6$

We know that

The fair-sided die has 6 possible outcomes. i.e.

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

Because each outcome is equally likely, we only have one chance in six to get a $6$on one roll.

As a result, the number of favorable outcomes is one, while the total number of possible outcomes is six.

The probability is calculated by dividing the number of favorable outcomes by the total number of possible outcomes.

$P\left(A\right)=\frac{\text{Number of favorable outcomes}}{\text{Numberof possible outcomes}}=\frac{1}{6}$

Apply the complement rule:

$P\left({\mathrm{A}}^{\mathrm{c}}\right)=1-P(\mathrm{A})=1-\frac{1}{6}=\frac{5}{6}$

Because the die has been rolled ten times, it's easier to assume that each roll is independent.

As a result, apply the multiplication rule for independent events if none of the ten rolls result in a $6$

$$\begin{gathered} P\left(B^c\right)=P\left(A^c\right)\times P\left(A^c\right).....\times P\left(A^c\right)=(P{\left(A^c\right))}^{10}={\left(\frac{5}{5}\right)}^{10}=\frac{9,765,625}{60,466,176} \\ \approx 0.1615 \end{gathered}$$

Apply the complement rule:

$P(\mathrm{B})=P\left({\left({\mathrm{B}}^{\mathrm{c}}\right)}^{\mathrm{c}}\right)=1-P\left({\mathrm{B}}^{\mathrm{c}}\right)=1-0.1615=0.8385$

Therefore, the Probability that at least $1$of the $10$rolls result in a $6$is $0.8385$