Answer to Question #132042 in Statistics and Probability for hasan

Question

Let X have a binomial distribution with n = 4 and P = 1/3 Find the probability. (i) P (X = 1) (ii) P (X = 3/2 ) (iii) P (X = 3) (iv) P (X = 6) (v) P (X ≤ 2) *

Answer 1

$P(X=k) = ( \begin{matrix} n \\ k \end{matrix}) p^k(1-p)^{n-k}$ - binomial distribution that describes k successes in n independent experiments. k=0,1,...,n. In our case: p =1/3, q=2/3, n=4, k={0,1,2,3,4}.

(i)$P (X = 1) = \frac{4!}{1!\; 3!} (\frac{1}{3})(\frac{2}{3})^3 = 4 \cdot \frac{8}{81} = \frac{32}{81} = 0.395$

(ii) $P(X=\frac{3}{2}) = 0$ since $X=\frac{3}{2}$ is not integer

(iii) $P (X = 3)= \frac{4!}{3!\; 1!} (\frac{1}{3})^3 \frac{2}{3} = 4 \cdot \frac{2}{81} = \frac{8}{81} = 0.0987$

(iv) $P (X = 6) = 0$, because possible value of k is grater than n, we cannot get more successful results that amount of experiments.

(v) $P (X ≤ 2) = P(X=0)+ P(X=1) + P(X=2)$

$P(X=0) = \frac{4!}{0!\; 4!} (\frac{1}{3})^0 (\frac{2}{3})^4= 1 \cdot \frac{16}{81} = 0.197$

$P(X=2) = \frac{4!}{2!\; 2!} (\frac{1}{3})^2 (\frac{2}{3})^2 = \frac{4 \cdot 3}{2} \cdot \frac{1}{9} \cdot \frac{4}{9} = \frac{24}{81} = 0.296$

$P (X ≤ 2) = 0.197+0.395+0.296=0.888$