Answer to Question #180663 in Statistics and Probability for EUGINE HAWEZA

Question

The incidence of COVID 19-disease is such that on the average twenty percent people suffer from it. If 100 people are selected at random, what is the chance that not more than ten people suffer from COVID 19-disease

Answer 1

p = 0.20

q= 1-0.20 =0.80

n = 100

X~B(n=100,p=0.20)

$P(X≤10) = P(X=0) + P(X=1) + P(X=2) + P(X=3) +P(X=4) + P(X=5) +P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(X=10) \\ P(X=x) = \binom{100}{x}p^xq^{100-x} \\ P(X=0) = \binom{100}{0}p^0q^{100-0} \\ = 1 \times 1 \times 0.8^{100} \\ = 2.037\times 10^{-10} \\ P(X=1) = \binom{100}{1}p^1q^{100-1} \\ = \frac{100!}{1!(100-1)!} \times 0.2^1 \times 0.8^{100-1} \\ = 100 \times 0.2 \times 0.8^{99} \\ =5.092\times 10^{-9} \\ P(X=2) = \binom{100}{2}p^2q^{100-2} \\ = \frac{100!}{2!(100-2)!} \times 0.2^2 \times 0.8^{100-2} \\ = \frac{99 \times 100}{2} \times 0.2^2 \times 0.8^{98} \\ = 6.302\times 10^{-8} \\ P(X=3) = \binom{100}{3}p^3q^{100-3} \\ = \frac{100!}{3!(100-3)!} \times 0.2^3 \times 0.8^{100-3} \\ = \frac{98 \times 99 \times 100}{2\times 3} \times 0.2^3 \times 0.8^{97} \\ = 5.146\times 10^{-7} \\ P(X=4) = \binom{100}{4}p^4q^{100-4} \\ = \frac{100!}{4!(100-4)!} \times 0.2^4 \times 0.8^{100-4} \\ = \frac{97 \times 98 \times 99 \times 100}{2\times 3 \times 4} \times 0.2^4 \times 0.8^{96} \\ = 3.120\times 10^{-6} \\ P(X=5) = \binom{100}{5}p^5q^{100-5} \\ = \frac{100!}{5!(100-5)!} \times 0.2^5 \times 0.8^{100-5} \\ = \frac{96 \times 97 \times 98 \times 99 \times 100}{2\times 3 \times 4 \times 5} \times 0.2^5 \times 0.8^{95} \\ =1.497 \times 10^{-5} \\ P(X=6) = \binom{100}{6}p^6q^{100-6} \\ = \frac{100!}{6!(100-6)!} \times 0.2^6 \times 0.8^{100-6} \\ = \frac{95 \times 96 \times 97 \times 98 \times 99 \times 100}{2\times 3 \times 4 \times 5 \times 6} \times 0.2^6 \times 0.8^{94} \\ = 2.928 \times 10^{-5} \\ P(X=7) = \binom{100}{7}p^7q^{100-7} \\ = \frac{100!}{7!(100-7)!} \times 0.2^7 \times 0.8^{100-7} \\ = \frac{94 \times 95 \times 96 \times 97 \times 98 \times 99 \times 100}{2\times 3 \times 4 \times 5 \times 6 \times 7} \times 0.2^7 \times 0.8^{93} \\ = 0.000199 \\ P(X=8) = \binom{100}{8}p^8q^{100-8} \\ = \frac{100!}{8!(100-8)!} \times 0.2^8 \times 0.8^{100-8} \\ = \frac{93 \times 94 \times 95 \times 96 \times 97 \times 98 \times 99 \times 100}{2\times 3 \times 4 \times 5 \times 6 \times 7 \times 8} \times 0.2^8 \times 0.8^{92} \\ = 0.000578 \\ P(X=9) = \binom{100}{9}p^9q^{100-9} \\ = \frac{100!}{9!(100-9)!} \times 0.2^9 \times 0.8^{100-9} \\ = \frac{92 \times 93 \times 94 \times 95 \times 96 \times 97 \times 98 \times 99 \times 100}{2\times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9} \times 0.2^9 \times 0.8^{91} \\ = 0.001478 \\ P(X=10) = \binom{100}{10}p^10q^{100-10} \\ = \frac{100!}{10!(100-10)!} \times 0.2^{10} \times 0.8^{100-10} \\ = \frac{91 \times 92 \times 93 \times 94 \times 95 \times 96 \times 97 \times 98 \times 99 \times 100}{2\times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10} \times 0.2^{10} \times 0.8^{90} \\ = 0.003362 \\ P(X≤10) = 0.005696$

0.57 % is the chance that not more than ten people suffer from COVID 19-disease.