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)=(20!)/((20-0)!0!)=(20!)/(20!*1)=1 \\ (20binomial1)=(20!)/(19!1!)=20 \\ (20binomial2)=(20!)/(18!2!)=(20*19)/(2)=190 \\ (20binomial3)=(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