Answer to Question #307387 in Statistics and Probability for hamii

Question

A coin is biased such that it is 1:2 for heads. If the coin is tossed 6 times, find the probabilities of obtaining various number of heads. 

Answer 1

The probability that we get head $p = \frac{2}{3},$ that we get tail $q = 1-p=1-\frac{2}{3}=\frac{1}{3}.$

Obviously the number of heads X we get after 6 tosses may be any value of 0, 1, 2, 3, 4, 5, 6.

We have a Bernoulli trial - exactly two possible outcomes, "success" (we get head) and "failure" (tail) and the probability of success is the same every time the experiment is conducted (the coin is tossed).

The pobability of each result

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

$=\cfrac{n!}{k!\cdot{(n-k)!}} \cdot \begin{pmatrix} \cfrac{2}{3} \end{pmatrix}^k \cdot \begin{pmatrix} \cfrac{1}{3} \end{pmatrix}^{n-k};$

$P(X=0)=\cfrac{6!}{0!\cdot{6!}} \cdot \begin{pmatrix} \cfrac{2}{3} \end{pmatrix}^0 \cdot \begin{pmatrix} \cfrac{1}{3} \end{pmatrix}^{6} =\cfrac{1}{3^6}=\cfrac{1}{729};$

$P(X=1)=\cfrac{6!}{1!\cdot{5!}} \cdot \begin{pmatrix} \cfrac{2}{3} \end{pmatrix}^1 \cdot \begin{pmatrix} \cfrac{1}{3} \end{pmatrix}^{5} =\cfrac{6\cdot2}{3^6}=\cfrac{12}{729};$

$P(X=2)=\cfrac{6!}{2!\cdot{4!}} \cdot \begin{pmatrix} \cfrac{2}{3} \end{pmatrix}^2 \cdot \begin{pmatrix} \cfrac{1}{3} \end{pmatrix}^{4} =\cfrac{5\cdot6\cdot2^2}{2\cdot3^6}=\cfrac{60}{729};$

$P(X=3)=\cfrac{6!}{3!\cdot{3!}} \cdot \begin{pmatrix} \cfrac{2}{3} \end{pmatrix}^3 \cdot \begin{pmatrix} \cfrac{1}{3} \end{pmatrix}^{3} =\cfrac{4\cdot5\cdot6\cdot2^3}{2\cdot3\cdot3^6}=\cfrac{160}{729};$

$P(X=4)=\cfrac{6!}{4!\cdot{2!}} \cdot \begin{pmatrix} \cfrac{2}{3} \end{pmatrix}^4 \cdot \begin{pmatrix} \cfrac{1}{3} \end{pmatrix}^{2} =\cfrac{5\cdot6\cdot2^4}{2\cdot3^6}=\cfrac{240}{729};$

$P(X=5)=\cfrac{6!}{5!\cdot{1!}} \cdot \begin{pmatrix} \cfrac{2}{3} \end{pmatrix}^5 \cdot \begin{pmatrix} \cfrac{1}{3} \end{pmatrix}^{1} =\cfrac{6\cdot2^5}{3^6}=\cfrac{192}{729};$

$P(X=6)=\cfrac{6!}{6!\cdot{0!}} \cdot \begin{pmatrix} \cfrac{2}{3} \end{pmatrix}^6 \cdot \begin{pmatrix} \cfrac{1}{3} \end{pmatrix}^{0} =\cfrac{2^6}{3^6}=\cfrac{64}{729}.$