Answer to Question #338666 in Statistics and Probability for sadiya

Question

On an average 5% of the population in an area suffers from Diabetes. What is the probability that out of 7 persons chosen at random from this area with less than and equals to three suffer from the disease

Answer 1

Let $X=$ the number of persons suffer from the disease :$X\sim Bin(n, p).$

Given $n=7 p=0.05, q=1-p=0.95.$

$$P(X\le)=P(X=0)+P(X=1)$$

$$+P(X=2)+P(X=3)$$

$$=\dbinom{7}{0}(0.05)^{0}(0.95)^{7-0}+\dbinom{7}{1}(0.05)^{1}(0.95)^{7-1}$$

$$\approx0.0002927$$

$$+\dbinom{7}{2}(0.05)^{2}(0.95)^{7-2}+\dbinom{7}{3}(0.05)^{3}(0.95)^{7-3}$$

$$=0.99980642188$$