ii) exactly 15 are defective

p(15,100,0.20) = $\displaystyle \frac{100!}{15! . {\color{blue}95}!} . (0.20)^{15} . ({\color{blue}0.20})^{{\color{blue}95}}$ = 0.005

Mr F says: Totally wrong. There are several basic mistakes (highlighted in blue).
i) at most 15 are defective ................

Question 1
how should i do this ...........

P(x $\displaystyle \le$ 15,100,0.20) = $\displaystyle \sum_{x = {\color{blue}1}}^{15} \frac{100!}{x! . (100 - x)!} (0.20)^x . (1-0.20)^{100-x}$ ........

Is this correct ........... Mr F says: Not quite. The sum should start from x = 0. Question 2) And fr Normal Approximation should i use the following formula below
z = $\displaystyle \frac{X - n \theta}{\sqrt{n \theta (1 - \theta)}}$

Mr F says: Yes. But do to the calculation you will need to use the continuity correction.