Question

I’m having hard time doing probality 3 most successes n=20 p=0.05

Answer 1

The binomial distribution of probability is

$\begin{gathered} P(X=k)=(nbinomialk)p^k(1-p)^{n-k} \\ n\rightarrow\text{ total number of trials} \\ k\rightarrow\text{ number of successful trials} \\ p\rightarrow\text{ probability of success} \end{gathered}$

Then, in our case,

$P(X=k)=(20binomialk)(0.05)^k(0.95)^{20-k}$

Therefore,

$\begin{gathered} \Rightarrow P(X\leq3)=P(X=0)+P(X=1)+P(X=2)+P(X=3) \\ \Rightarrow P(X\leq3)=(20binomial0)(0.05)^0(0.95)^{20}+(20binomial1)(0.05)^1(0.95)^{19}+... \\ ...+(20binomial2)(0.05)^2(0.95)^{18}+(20binomial3)(0.05)^3(0.95)^{17} \end{gathered}$

Then,

$\begin{gathered} (20binomial0)=\frac{20!}{(20-0)!0!}=\frac{20!}{20!*1}=1 \\ (20binomial1)=\frac{20!}{19!1!}=20 \\ (20binomial2)=\frac{20!}{18!2!}=\frac{20*19}{2}=190 \\ (20binomial3)=\frac{20!}{17!3!}=1140 \end{gathered}$

Thus, after substituting the values of the binomial coefficients,

$\Rightarrow P(X\leq3)\approx0.9841...$

The answer is approximately P(X<=3)=0.9841