Question

Renzo tossed a pair of dice five times. What is the probability that he will get at least three "9"

Answer 1

The total number of possible outcomes for a single toss of a pair of dice $N=6\cdot6=36.$

The favorable outcomes (Renzo gets 9) are (3, 6), (4, 5), (5, 4), (6, 3); $n=4.$

The probability of getting a 9 for each toss is $p=\cfrac{4}{36}=\cfrac{1}{9}.$

The probability of getting another sum is $q=1-\cfrac{1}{9}=\cfrac{8}{9}.$

We have a Bernoulli trial - exactly two possible outcomes, "success" (Renzo gets a 9) and "failure" (Renzo gets another sum) and the probability of success is the same every time the experiment is conducted (Renzo tosses a pair of dice).

The probability of each result

$P(X=k)=\begin{pmatrix}n\\k\end{pmatrix}\cdot p^k\cdot q^{n-k}=\\ =\begin{pmatrix}5\\k\end{pmatrix}\cdot \begin{pmatrix}\cfrac{1}{9}\end{pmatrix}^{k}\cdot \begin{pmatrix}\cfrac{8}{9}\end{pmatrix}^{5-k}=\\ =\cfrac{5!}{k!\cdot(5-k)!}\cdot \begin{pmatrix}\cfrac{1}{9}\end{pmatrix}^{k}\cdot \begin{pmatrix}\cfrac{8}{9}\end{pmatrix}^{5-k}\\.$

The sought probability that Renzo gets at least three 9:

$P(X\geq 3)=\\ =P(X=3)+P(X=4)+P(X=5)=\\ =\cfrac{5!}{3!\cdot2!}\cdot \begin{pmatrix}\cfrac{1}{9}\end{pmatrix}^{3}\cdot \begin{pmatrix}\cfrac{8}{9}\end{pmatrix}^2+\cfrac{5!}{4!\cdot1!}\cdot \begin{pmatrix}\cfrac{1}{9}\end{pmatrix}^{4}\cdot \begin{pmatrix}\cfrac{8}{9}\end{pmatrix}^1+\\ +\cfrac{5!}{5!\cdot0!}\cdot \begin{pmatrix}\cfrac{1}{9}\end{pmatrix}^{5}\cdot \begin{pmatrix}\cfrac{8}{9}\end{pmatrix}^0=\\ =0,011533.$