Answer to Question #341127 in Statistics and Probability for Mohanad

Question

the probability of winning is 1/3. Mona plays 6 games. Find probability of winning exactly 5 games.

Answer 1

We have a Bernoulli trial - exactly two possible outcomes, "success" (Mona wins the game) and "failure" (Mona doesn't win the game) and the probability of success is the same every time the experiment is conducted (Mona plays a game), $p=\cfrac{1}{3}, q=1-\cfrac{1}{3}=\cfrac{2}{3}, n=6.$

The probability that Mona wins k games

$P(X=k)=\begin{pmatrix}n\\k\end{pmatrix}\cdot p^k\cdot q^{n-k}=\\ =\begin{pmatrix}6\\k\end{pmatrix}\cdot \begin{pmatrix}\cfrac{1}{3}\end{pmatrix} ^k\cdot \begin{pmatrix}\cfrac{2}{3}\end{pmatrix}^{6-k}=\\ =\cfrac{6!}{k!\cdot(6-k)!}\cdot \begin{pmatrix}\cfrac{1}{3}\end{pmatrix}^k\cdot \begin{pmatrix}\cfrac{2}{3}\end{pmatrix}^{6-k}.$

The probability that Mona wins 5 games

$P(X=5)=\cfrac{6!}{5!\cdot1!}\cdot \begin{pmatrix}\cfrac{1}{3}\end{pmatrix}^{5}\cdot \begin{pmatrix}\cfrac{2}{3}\end{pmatrix}^{1}=0.0165.$